Chapter 2 part 2 : Force and Law of motion class 9 solution
Very Short Answer Type Questions
Answers to Physics Questions
1. Which physical quantity corresponds to the rate of change of momentum?
Answer: Force.
2. State the relation between the momentum of a body and the force acting on it.
Answer: According to Newton's Second Law of Motion, the rate of change of momentum of a body is directly proportional to the force acting on it and occurs in the same direction as the force. Mathematically, F = dp/dt, where F is the force, p is the momentum, and t is time.
3. What is the unit of force?
Answer: The SI unit of force is the newton (N).
4. Define one newton force.
Answer: One newton is the force required to produce an acceleration of 1 meter per second squared in a mass of 1 kilogram. It can be represented as 1 N = 1 kg * m/s².
5. What is the relationship between force and acceleration?
Answer: According to Newton's Second Law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, F = ma, where F is the net force, m is the mass, and a is the acceleration.
6. If the mass of a body and the force acting on it are both doubled, what happens to the acceleration?
Answer: Doubling both the mass and the force acting on it will result in the acceleration remaining the same. This is because the increase in force is compensated for by the increase in mass.
7. Name the physical quantity whose unit is ‘newton’.
Answer: Force.
8. Which physical principle is involved in the working of a jet aeroplane?
Answer: Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. The jet engine generates thrust by pushing hot air backwards, and the reaction propels the airplane forward.
9. Name the principle on which a rocket works.
Answer: Newton's Third Law of Motion. The rocket burns fuel, which creates hot gases that are expelled from the rocket nozzle. This creates a forward thrust that propels the rocket in the opposite direction.
10. Is the following statement true or false: A rocket can propel itself in a vacuum.
Answer: True. A rocket can propel itself in a vacuum because it carries its own fuel and oxidizer, which it burns to create thrust. It does not rely on the air around it to provide a pushing force.
11. What is the force which produces an acceleration of 1 m/s2 in a body of mass 1 kg?
Answer: According to F = ma, the force required to produce an acceleration of 1 m/s2 in a body of mass 1 kg is:
F = 1 kg * 1 m/s² = 1 N.
12. Find the acceleration produced by a force of 5 N acting on a mass of 10 kg.
Answer: Using F = ma, the acceleration produced by the force is:
a = F / m = 5 N / 10 kg = 0.5 m/s².
13. A girl weighing 25 kg stands on the floor. She exerts a downward force of 250 N on the floor. What force does the floor exert on her?
Answer: According to Newton's Third Law, the force exerted by the floor on the girl is equal and opposite to the force she exerts on the floor, which is 250 N upwards.
14. Name the physical quantity which makes it easier to accelerate a small car than a large car.
Answer: Mass. Since acceleration is inversely proportional to mass, a smaller car with less mass will experience a greater acceleration with the same force compared to a larger car with more mass.
15. Fill in the following blanks with suitable words :
(a) To every action, there is an equal and opposite reaction.
(b) Momentum is a vector quantity. Its unit is kilogram meters per second.
(c) Newton’s second law of motion can be written as Force = mass × acceleration or Force = rate of change of momentum.
(d) Forces in a Newton’s third law pair have equal magnitude but act in opposite directions.
(e) In collisions and explosions, the total momentum remains constant, provided that no external force acts.
Short Answer Type Questions
16. Explain the meaning of the following equation : F = m × a where symbols have their usual meanings.
Answer: This equation represents Newton's Second Law of Motion. It states that the net force (F) acting on an object is directly proportional to the mass (m) of the object and its acceleration (a).
Explanation:
- Force (F): This refers to the push or pull exerted on an object, measured in Newtons (N).
- Mass (m): This is the quantity of matter in the object, measured in kilograms (kg).
- Acceleration (a): This is the rate of change of velocity of the object, measured in meters per second squared (m/s²).
Therefore, the equation tells us that the greater the force applied to an object, the greater its acceleration will be, assuming its mass remains constant. Similarly, the greater the mass of an object, the greater the force required to achieve the same acceleration.
17. To take the boat away from the bank of a river, the boatman pushes the bank with an oar. Why?
Answer: The boatman pushes the bank with his oar to create a force (F) against the bank. According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means the bank pushes back on the boat with an equal and opposite force (F), propelling it away from the shore.
18. Why does a gunman get a jerk on firing a bullet?
Answer: When a gunman fires a bullet, the explosion of the gunpowder propels the bullet forward with a force (F). According to Newton's Third Law, the gun experiences an equal and opposite force (F) in the backward direction. This sudden backward force causes the gunman to experience a jerk.
19. If action is always equal to reaction, explain why a cart pulled by a horse can be moved.
Answer: While action and reaction are equal and opposite, they do not necessarily act on the same object. In the case of a horse pulling a cart:
- The horse exerts a force (F) forward on the cart.
- The cart, in turn, exerts an equal and opposite force (F) backward on the horse.
However, the cart is also subject to the frictional force of the ground, which opposes the forward motion. If the horse exerts a force greater than the frictional force, the net force on the cart will be positive, causing it to move forward.
Therefore, even though the horse and the cart experience equal and opposite forces, the cart moves forward because of the imbalance in the overall forces acting on it. The horse's forward force overcomes the opposing forces, resulting in the cart's movement.
20. Explain how a rocket works.
Answer: A rocket works on the principle of Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction. The rocket carries its own fuel and oxidizer, which it burns to create hot gases. These gases are then expelled from the rocket engine through a nozzle at high velocity.
This rapid expulsion of gas creates a forward thrust that propels the rocket in the opposite direction. As the rocket burns more fuel and expels more gas, it continues to accelerate. This process allows rockets to launch into space and travel through the vacuum, where there is no air to provide lift or thrust.
21. Do action and reaction act on the same body or different bodies? How are they related in magnitude and direction? Are they simultaneous or not?
Answer: Action and reaction always act on different bodies. They are equal in magnitude but opposite in direction. They are also simultaneous, meaning they occur at the same time.
For example, when you push a wall with your hand (action), the wall pushes back on your hand with an equal force in the opposite direction (reaction). Both forces are equal and opposite, and they occur at the same instant.
22. If a man jumps out from a boat, the boat moves backwards. Why?
Answer: This is again due to Newton's Third Law of Motion. When the man jumps out of the boat, he exerts a force on the boat (action) by pushing off with his feet. The boat, in turn, exerts an equal and opposite force on the man (reaction), propelling him forward. However, since the man's mass is much smaller than the boat's mass, his acceleration will be much greater. As a result, the boat will experience a slight backward movement.
23. Why is it difficult to walk on a slippery road?
Answer: Walking on a slippery road is difficult because the friction between your shoes and the ground is reduced. Friction is the force that helps us push against the ground and move forward. When the surface is slippery, there is less friction, making it harder for us to grip the ground and propel ourselves forward. This can result in slipping and falling.
24. Explain why a runner presses the ground with his feet before he starts his run.
Answer: By pressing the ground hard with his feet before starting his run, the runner increases the friction between his shoes and the ground. This provides him with more traction and allows him to push off with greater force, resulting in a faster start and more efficient running.
25. A 60 g bullet fired from a 5 kg gun leaves with a speed of 500 m/s. Find the speed (velocity) with which the
gun recoils (jerks backwards).
Solution:
Given:
- Mass of the bullet (m_bullet) = 60 g = 0.06 kg
- Mass of the gun (m_gun) = 5 kg
- Speed of the bullet (v_bullet) = 500 m/s
- Speed of the gun (v_gun) = unknown
Using the principle of conservation of momentum:
Total momentum before firing = Total momentum after firing
Before firing:
- Momentum of the bullet (p_bullet) = m_bullet * v_bullet = 0.06 kg * 500 m/s = 30 kg m/s
- Momentum of the gun (p_gun) = m_gun * v_gun = 5 kg * v_gun
After firing:
- Momentum of the bullet (p_bullet) = 30 kg m/s (remains the same)
- Momentum of the gun (p_gun) = 5 kg * (-v_gun) (negative sign indicates opposite direction)
Equating momentum before and after firing:
30 kg m/s + 5 kg * v_gun = 0.06 kg * 500 m/s
Solving for v_gun:
v_gun = -30 kg m/s / 5 kg = -6 m/s
Therefore, the speed with which the gun recoils is 6 m/s (backward).
26. A 10 g bullet travelling at 200 m/s strikes and remains embedded in a 2 kg target which is originally at rest
but free to move. At what speed does the target move off ?
Solution:
Given:
- Mass of bullet (m_b) = 10 g = 0.01 kg
- Initial velocity of bullet (v_b) = 200 m/s
- Mass of target (m_t) = 2 kg
- Initial velocity of target (v_t) = 0 m/s (at rest)
Unknown:
- Final velocity of target (v_f)
Formula:
We can use the law of conservation of momentum:
m_b * v_b + m_t * v_t = (m_b + m_t) * v_f
Solving for v_f:
- Substitute the given values:
0.01 kg * 200 m/s + 2 kg * 0 m/s = (0.01 kg + 2 kg) * v_f
- Rearrange the equation to solve for v_f:
v_f = (0.01 kg * 200 m/s) / (0.01 kg + 2 kg)
- Calculate the final velocity:
v_f = 0.99 m/s
Therefore, the target moves off at a speed of 0.99 m/s.
27. A body of mass 2 kg is at rest. What should be the magnitude of force which will make the body move with
a speed of 30 m/s at the end of 1 s ?
Solution:
Given:
- Mass of the body (m) = 2 kg
- Initial velocity (u) = 0 m/s (at rest)
- Final velocity (v) = 30 m/s
- Time interval (t) = 1 s
Unknown:
- Force (F)
Formula:
We can use the equation of motion:
v = u + at
where a is the acceleration of the body.
Solving for a:
- Rearrange the equation to isolate a:
a = (v - u) / t
- Substitute the given values:
a = (30 m/s - 0 m/s) / 1 s
- Calculate the acceleration:
a = 30 m/s²
Solving for F:
- Use Newton's Second Law:
F = ma
- Substitute the values of mass and acceleration:
F = 2 kg * 30 m/s²
- Calculate the force:
F = 60 N
Therefore, the magnitude of the force required to make the body move with a speed of 30 m/s at the end of 1 s is 60 N.
28. A body of mass 5 kg is moving with a velocity of 10 m/s. A force is applied to it so that in 25 seconds, it
attains a velocity of 35 m/s. Calculate the value of the force applied.
Solution:
Given:
- Mass of body (m) = 5 kg
- Initial velocity (u) = 10 m/s
- Final velocity (v) = 35 m/s
- Time interval (t) = 25 s
Unknown:
- Force applied (F)
Formula:
We can use the equation of motion:
v = u + at
where a is the acceleration of the body.
Solving for a:
- Rearrange the equation to isolate a:
a = (v - u) / t
- Substitute the given values:
a = (35 m/s - 10 m/s) / 25 s
- Calculate the acceleration:
a = 1 m/s²
Solving for F:
- Use Newton's Second Law:
F = ma
- Substitute the values of mass and acceleration:
F = 5 kg * 1 m/s²
- Calculate the force:
F = 5 N
Therefore, the value of the force applied to the body is 5 N.
29. A car of mass 2400 kg moving with a velocity of 20 m s–1 is stopped in 10 seconds on applying brakes.
Calculate the retardation and the retarding force.
Solution:
Given:
- Mass of car (m) = 2400 kg
- Initial velocity (u) = 20 m/s
- Final velocity (v) = 0 m/s (stopped)
- Time interval (t) = 10 s
Unknown:
- Retardation (a)
- Retarding force (F)
Solving for retardation:
- Use the equation of motion:
v = u + at
- Rearrange the equation to isolate a:
a = (v - u) / t
- Substitute the given values:
a = (0 m/s - 20 m/s) / 10 s
- Calculate the retardation:
a = -2 m/s² (negative sign indicates deceleration)
Solving for retarding force:
- Use Newton's Second Law:
F = ma
- Substitute the values of mass and acceleration:
F = 2400 kg * (-2 m/s²)
- Calculate the retarding force:
F = -4800 N (negative sign indicates opposing force)
Therefore:
- The retardation of the car is 2 m/s².
- The retarding force applied by the brakes is 4800 N.
30. For how long should a force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 m/s ?
Solution:
Given:
- Force applied (F) = 100 N
- Mass of body (m) = 20 kg
- Final velocity (v) = 100 m/s
- Initial velocity (u) assumed to be 0 m/s (at rest)
Unknown:
- Time interval (t)
Formula:
We can use the equation of motion:
v = u + at
Solving for a:
- Rearrange the equation to isolate a:
a = (v - u) / t
- Substitute the given values:
a = (100 m/s - 0 m/s) / t
- Calculate the acceleration:
a = 100 m/s²
Solving for t:
- Use Newton's Second Law:
F = ma
- Rearrange the equation to isolate t:
t = F / (ma)
- Substitute the given values:
t = 100 N / (20 kg * 100 m/s²)
- Calculate the time interval:
t = 0.05 s
Therefore, the force of 100 N should act on the body for 0.05 seconds to make it reach a velocity of 100 m/s.
31. How long will it take a force of 10 N to stop a mass of 2.5 kg which is moving at 20 m/s ?
Answer: The time it takes to stop the mass can be calculated using the equation , where is the time, is the mass, is the velocity, and is the force.
Plugging in the values:
So, it will take 5 seconds for the force of 10 N to stop the mass.
32. The velocity of a body of mass 10 kg increases from 4 m/s to 8 m/s when a force acts on it for 2 s. (a) What is the momentum before the force acts ? (b) What is the momentum after the force acts ? (c) What is the gain in momentum per second ? (d) What is the value of the force ?
Solution:
(a) Momentum before the force acts:
Momentum is defined as the mass of an object multiplied by its velocity.
Formula: p = mv
Given: m = 10 kg, u = 4 m/s (initial velocity)
Calculation: p₁ = 10 kg * 4 m/s = 40 kg⋅m/s
Therefore, the momentum before the force acts is 40 kg⋅m/s.
(b) Momentum after the force acts:
Given: v = 8 m/s (final velocity)
Calculation: p₂ = 10 kg * 8 m/s = 80 kg⋅m/s
Therefore, the momentum after the force acts is 80 kg⋅m/s.
(c) Gain in momentum per second:
Formula: Gain in momentum per second = Δp / Δt
Given: Δp = p₂ - p₁ = 80 kg⋅m/s - 40 kg⋅m/s = 40 kg⋅m/s Δt = 2 s (time interval)
Calculation: Gain in momentum per second = 40 kg⋅m/s / 2 s = 20 kg⋅m/s²
Therefore, the gain in momentum per second is 20 kg⋅m/s².
(d) Value of the force:
Formula: F = Δp / Δt = ma
Given: Δp = 40 kg⋅m/s, Δt = 2 s, m = 10 kg
Calculation: F = 40 kg⋅m/s / 2 s = 20 N
Therefore, the value of the force is 20 N.
33. A gun of mass 3 kg fires a bullet of mass 30 g. The bullet takes 0.003 s to move through the barrel of the gun
and acquires a velocity of 100 m/s. Calculate :
(i) the velocity with which the gun recoils.
(ii) the force exerted on gunman due to recoil of the gun
(i) The velocity with which the gun recoils.
Answer: According to the law of conservation of momentum, the momentum of the gun and bullet system before firing is equal to the momentum after firing.
Here, and are the mass and velocity of the gun before firing, and and are the mass and velocity of the bullet before firing.
Since the gun is initially at rest (), the equation becomes:
Solving for :
Substitute the given values:
So, the velocity with which the gun recoils is (negative sign indicates the opposite direction to the bullet).
(ii) The force exerted on the gunman due to the recoil of the gun.
Answer: The force exerted on the gunman can be calculated using Newton's second law:
Since acceleration () is the change in velocity over time:
Substitute the given values:
So, the force exerted on the gunman due to the recoil of the gun is approximately (negative sign indicates the opposite direction to the bullet).
34. Draw a diagram to show how a rocket engine provides a force to move the rocket upwards. Label the diagram appropriately.
Answer:
Rocket Engine Diagram with Labels:
/--------------------\
/ \
/ \
/ \
/ \
|------------------------|
| |
| |
| Fuel Tank (Oxidizer + |
| Fuel) |
| |
| |
|------------------------|
/|\ /|\ /|\
/ | \ / | \ / | \
Nozzle --> / | \ / | \ / | \
/ | \/ | \/ | \
/ | / | / |
/ | / | / |
/ | / | / |
/ | / | / |
/--------| /--------| /--------|
/ | / | / |
/ Combustion |/ Combustion |/ Combustion |
/ Chamber | Chamber | Chamber |
/ |/ |/ |
/ | | |
|-----------------/-----------------/-----------------|
| Exhaust | Exhaust | Exhaust |
| | | |
|--------------------------------------------------------------------|
| ^ |
| Upward Thrust (Force) |
|--------------------------------------------------------------------|
Label Descriptions:
- Fuel Tank: This container holds the fuel and oxidizer needed for combustion.
- Combustion Chamber: This chamber is where the fuel and oxidizer mix and burn, creating hot gases.
- Nozzle: This funnel-shaped part directs the hot gases out of the engine, creating thrust.
- Exhaust: The hot gases expelled from the nozzle, providing the force that propels the rocket forward.
- Upward Thrust (Force): The forward force generated by the expelled exhaust gases, pushing the rocket upwards.
Note: This is a simplified diagram, and the actual design of a rocket engine may vary depending on the type of engine.
35. Name the laws involved in the following situations :
(a) the sum of products of masses and velocities of two moving bodies before and after their collision remains
the same.
(b) a body of mass 5 kg can be accelerated more easily by a force than another body of mass 50 kg under
similar conditions
(c) when person A standing on roller skates pushes another person B (also standing on roller skates) and
makes him move to the right side, then the person A himself gets moved to the left side by an equal
distance.
(d) if there were no friction and no air resistance, then a moving bicycle would go on moving for ever.
- Laws involved in the following situations:
(a) Law of Conservation of Linear Momentum:
- This law states that the total linear momentum of a system of objects remains constant if no external forces act on the system.
- In the context of the given situation, the sum of the products of masses and velocities of two moving bodies before and after their collision remains the same. This is a manifestation of the conservation of linear momentum, where the total momentum of the system is conserved during a collision.
(b) Newton's Second Law of Motion:
- Newton's Second Law states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, it is expressed as .
- In the given scenario, a body of mass 5 kg can be accelerated more easily by a force than another body of mass 50 kg under similar conditions. This aligns with Newton's Second Law, indicating that the acceleration of an object is directly proportional to the force applied and inversely proportional to its mass.
(c) Newton's Third Law of Motion:
- Newton's Third Law states that for every action, there is an equal and opposite reaction.
- In the provided example, when person A pushes person B, causing person B to move to the right, person A experiences an equal and opposite force, causing them to move to the left. This is a direct application of Newton's Third Law, where the action and reaction forces are of equal magnitude but act in opposite directions.
(d) Law of Inertia or Newton's First Law of Motion:
- Newton's First Law states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity unless acted upon by a net external force.
- In the context of the moving bicycle, the statement reflects the idea that, in the absence of friction and air resistance (external forces), a body in motion (the bicycle) would continue moving indefinitely. This aligns with Newton's First Law, emphasizing the concept of inertia, which is the tendency of an object to maintain its state of motion.
- Long Answer Type Questions
- 36. (a) State and explain Newton’s second law of motion.
- (b) A 1000 kg vehicle moving with a speed of 20 m/s is
- brought to rest in a distance of 50 metres :
- (i) Find the acceleration.
- (ii) Calculate the unbalanced force acting on the vehicle.
(a) Newton's Second Law of Motion: Newton's second law of motion states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. Mathematically, it can be expressed as:
where:
- is the net force acting on the object (in Newtons),
- is the mass of the object (in kilograms),
- is the acceleration of the object (in meters per second squared).
This law essentially tells us how the motion of an object changes when it is subjected to an external force.
(b) Vehicle Stopping Distance: Given:
- Mass () = 1000 kg
- Initial speed () = 20 m/s
- Final speed () = 0 m/s (brought to rest)
- Distance () = 50 m
(i) Find the acceleration (): The equation relating initial velocity (), final velocity (), acceleration (), and distance () is given by:
In this case, since , the equation simplifies to:
Solving for :
(ii) Calculate the unbalanced force () acting on the vehicle: Now that we have the acceleration, we can use Newton's second law to find the force:
Substitute the values:
Solving for will give you the unbalanced force acting on the vehicle.
37. (a) Explain why, a cricket player moves his hands
backwards while catching a fast cricket ball.
(b) A 150 g ball, travelling at 30 m/s, strikes the palm of a
player’s hand and is stopped in 0.05 second. Find the
force exerted by the ball on the hand.
(a) Cricket player moving hands backwards while catching a fast ball
There are two primary reasons why a cricket player moves their hands backwards while catching a fast ball:
1. Reduce force: A fast-moving ball carries significant momentum. Stopping it abruptly requires a large force. By moving their hands backwards, the player increases the time interval over which the ball comes to rest. This increases the time it takes to change the ball's momentum, ultimately reducing the average force required to catch it.
Imagine pushing a heavy object. It's easier to push it slowly over a longer distance than to push it quickly over a shorter distance. The same principle applies to catching a ball.
2. Cushion the impact: Moving the hands back helps absorb the impact of the ball, reducing the stress on the player's hands and wrists. This is crucial for catching fast balls, which can be painful if caught directly.
Here's an analogy:
As you can see, the cricketer is moving their hands backwards as they catch the ball. This helps reduce the impact force and protect their hands.
(b) Force exerted by the ball on the hand
We can use the impulse-momentum theorem to solve this problem. This theorem states that the net impulse acting on an object is equal to the change in its momentum:
F * Δt = Δp
where:
- F is the net force acting on the object
- Δt is the time interval
- Δp is the change in momentum
In this scenario, the net force is the force exerted by the ball on the hand, and the change in momentum is the difference between the ball's momentum before and after being caught.
Given:
- Ball mass (m) = 150 g = 0.15 kg
- Initial ball velocity (u) = 30 m/s
- Final ball velocity (v) = 0 m/s
- Time interval (Δt) = 0.05 s
Change in momentum:
Δp = m * (v - u) Δp = 0.15 kg * (0 m/s - 30 m/s) Δp = -4.5 kg m/s
Force exerted by the ball on the hand:
F = Δp / Δt F = -4.5 kg m/s / 0.05 s F = -90 N
Therefore, the force exerted by the ball on the hand is 90 N.
38. (a) State Newton’s third law of motion and give two
examples to illustrate the law.
(b) Explain why, when a fireman directs a powerful stream
of water on a fire from a hose pipe, the hose pipe tends
to go backward.
(a) Newton's Third Law of Motion: Newton's third law of motion states that for every action, there is an equal and opposite reaction. This means that if object A exerts a force on object B, then object B simultaneously exerts a force of equal magnitude in the opposite direction on object A. Mathematically, it can be expressed as:
Here, is the force exerted by object A on object B, and is the force exerted by object B on object A.
Examples:
Walking: When you walk, your foot exerts a backward force on the ground (action), and the ground simultaneously exerts an equal and opposite forward force on your foot (reaction).
Swimming: As a swimmer pushes the water backward with their hands (action), the water exerts an equal and opposite force forward on the swimmer (reaction).
(b) Fireman and Hose Pipe: When a fireman directs a powerful stream of water from a hose pipe to extinguish a fire, the hose pipe tends to go backward due to Newton's third law. The explanation is based on the equal and opposite reaction forces.
As water is expelled from the hose at high speed in one direction (action), the hose experiences an equal and opposite force in the opposite direction (reaction). This is in accordance with Newton's third law. The force exerted on the water (action) results in an equal force acting on the hose pipe in the opposite direction (reaction). This phenomenon is known as the recoil effect. To minimize this effect, firefighters often brace themselves or use specialized equipment designed to handle the powerful force generated when water is expelled from the hose.
39. (a) State the law of conservation of momentum. (b) Discuss the conservation of momentum in each of the following cases : (i) a rocket taking off from ground. (ii) flying of a jet aeroplane.
(a) Law of Conservation of Momentum: The law of conservation of momentum states that the total momentum of an isolated system of objects remains constant if no external forces act on it. In other words, the momentum of a system before an event must be equal to the momentum after the event, provided no external forces are involved. Mathematically, the law can be expressed as:
(b) Conservation of Momentum in Specific Cases:
(i) Rocket Taking Off from Ground: When a rocket takes off from the ground, it expels gases at high speed in the downward direction. According to Newton's third law of motion, for every action, there is an equal and opposite reaction. The action is the expulsion of gases downward, and the reaction is the upward force on the rocket. As a result, the rocket gains momentum in the upward direction, while an equal amount of momentum is imparted to the expelled gases in the downward direction. Since there are no external horizontal forces, the horizontal component of momentum remains constant. The conservation of momentum is evident in the fact that the change in momentum of the rocket is equal and opposite to the change in momentum of the expelled gases.
(ii) Flying of a Jet Aeroplane: In the case of a jet aeroplane flying at a constant velocity, the principle of conservation of momentum can be applied. The engines of the jet expel air backward at a high speed, generating a forward thrust. According to Newton's third law, the backward expulsion of air is the action, and the forward thrust on the jet is the reaction. The forward momentum gained by the jet is equal and opposite to the backward momentum imparted to the expelled air. As a result, the total momentum of the system (jet + expelled air) remains constant. This illustrates the conservation of momentum during the flying of a jet aeroplane.
In both cases, the key is that the change in momentum of the object (rocket or jet) is matched by an equal and opposite change in momentum of the expelled mass (gases or air), leading to the overall conservation of momentum in the system.
40. (a) If a balloon filled with air and its mouth untied, is released with its mouth in the downward direction, it moves upwards. Why ? (b) An unloaded truck weighing 2000 kg has a maximum acceleration of 0.5 m/s2. What is the maximum acceleration when it is carrying a load of 2000 kg ?
(a) Balloon moving upwards
A balloon filled with air and released with its mouth downwards will move upwards for two main reasons:
1. Newton's third law of motion: As the air inside the balloon rushes out through the opening, it exerts a force downwards (action). According to Newton's third law, an equal and opposite force is exerted upwards on the balloon (reaction). This upward force propels the balloon upwards, overcoming the gravitational pull and causing it to rise.
2. Buoyancy of air: While the escaping air is propelling the balloon upwards, the air surrounding the balloon also exerts a buoyant force that counteracts gravity. This buoyant force is dependent on the difference in densities between the air inside and outside the balloon. As the density of air inside the balloon is higher due to inflation, the buoyant force pushes the balloon upwards further.
Therefore, the combination of upward thrust due to escaping air and the buoyant force of the surrounding air causes the balloon to move upwards even when released with its mouth downwards.
(b) Truck acceleration with and without load
The maximum acceleration of a vehicle depends on the net force acting on it and its mass. The net force is the difference between the driving force and the opposing forces (friction, air resistance, etc.).
Given:
- Unloaded truck weight (m1) = 2000 kg
- Maximum acceleration without load (a1) = 0.5 m/s²
- Load weight (m2) = 2000 kg
Truck mass with load (m_total) = m1 + m2 = 4000 kg
Assuming the driving force remains constant:
- Net force without load (F_net1) = m1 * a1
- Net force with load (F_net2) = m_total * a2
Since the driving force remains constant, F_net1 = F_net2
Therefore, m1 * a1 = m_total * a2
Solving for a2:
a2 = (m1 * a1) / m_total
a2 = (2000 kg * 0.5 m/s²) / 4000 kg
a2 = 0.25 m/s²
Therefore, the maximum acceleration of the truck with a 2000 kg load is 0.25 m/s².
This highlights that increasing the mass of the vehicle while keeping the driving force constant will decrease its maximum acceleration.
Multiple Choice Questions (MCQs)
41. The rockets work on the principle of conservation of :
(a) mass (b) energy (c) momentum (d) velocity
(c) momentum: Rockets work on the principle of conservation of momentum. The expulsion of mass in one direction (action) generates an equal and opposite momentum in the opposite direction (reaction), propelling the rocket forward.
42. An object of mass 2 kg is sliding with a constant velocity of 4 m/s on a frictionless horizontal table. The
force required to keep this object moving with the same velocity is :
(a) 32 N (b) 0 N (c) 2 N (d) 8 N
(b) 0 N: Since the object is sliding with a constant velocity on a frictionless surface, there is no net force required to keep it moving at a constant velocity. The force needed to overcome friction is zero.
43. The physical quantity which makes it easier to accelerate a small car than a large car is measured in the
unit of :
(a) m/s (b) kg (c) kg.m/s (d) kg.m/s2
(c) kg.m/s: The physical quantity that makes it easier to accelerate a small car than a large car is inertia. Inertia is measured by the mass of an object. The unit of mass is kg, and the unit of velocity is m/s, so the unit of the product (mass × velocity) is kg.m/s.
44. According to the third law of motion, action and reaction :
(a) always act on the same body but in opposite directions
(b) always act on different bodies in opposite directions
(c) have same magnitudes and directions
(d) act on either body at normal to each other
(b) always act on different bodies in opposite directions: According to Newton's third law of motion, action and reaction forces always act on different bodies in opposite directions.
45. The unit of measuring momentum of a moving body is :
(a) m s–1 (b) kg.m s–1 (c) kg.m s–2 (d) Nm2 kg–2
(b) kg.m s–1: The unit of measuring momentum is the product of mass and velocity. Therefore, the unit of momentum is kg.m/s.
46. A boy of mass 50 kg standing on ground exerts a force of 500 N on the ground. The force exerted by the
ground on the boy will be :
(a) 50 N (b) 25000 N (c) 10 N (d) 500 N
(d) 500 N: According to Newton's third law, the force exerted by the ground on the boy is equal in magnitude and opposite in direction to the force exerted by the boy on the ground. Therefore, the force exerted by the ground on the boy is also 500 N.
47. A Honda City car, a Maruti Alto car, a Tata Nano car and a Mahindra Scorpio car, all are running at the same speed of 50 m/s under identical conditions. If all these cars are hit from behind with the same force and they continue to move forward, the maximum acceleration will be produced in : (a) Honda City (b) Maruti Alto (c) Tata Nano (d) Mahindra Scorpio
The maximum acceleration produced is given by the formula , where is the force and is the mass. Since all the cars experience the same force, the one with the least mass will have the maximum acceleration. Therefore, the Tata Nano, being the lightest among the given cars, will have the maximum acceleration. So, the answer is (c) Tata Nano.
48. The acceleration produced by a force of 5 N acting on a mass of 20 kg in m/s2 is :
(a) 4 (b) 100 (c) 0.25 (d) 2.5
The formula for acceleration is , where is the force and is the mass. Given and , . So, the answer is (c) 0.25.
49. Which of the following situations involves the Newton’s second law of motion ?
(a) a force can stop a lighter vehicle as well as a heavier vehicle which are moving
(b) a force can accelerate a lighter vehicle more easily than a heavier vehicle which are moving
(c) a force exerted by a lighter vehicle on collision with a heavier vehicle results in both the vehicles coming
to a standstill
(d) a force exerted by the escaping air from a balloon in the downward direction makes the balloon to go
upwards
Newton's second law of motion is , where is the force applied, is the mass, and is the acceleration. Among the given options, (b) "a force can accelerate a lighter vehicle more easily than a heavier vehicle which are moving" involves Newton's second law.
50. A fielder pulls his hands backwards after catching the cricket ball. This enables the fielder to :
(a) exert larger force on the ball (b) reduce the force exerted by the ball
(c) increase the rate of change of momentum (d) keep the ball in hands firmly
When the fielder pulls his hands backward after catching the cricket ball, he increases the time of contact. This action increases the time over which the change in momentum occurs, reducing the force exerted by the ball on the hands. So, the answer is (b) reduce the force exerted by the ball.
Questions Based on High Order Thinking Skills (HOTS)
51. Why are car seat-belts designed to stretch somewhat in a collision ?
Car seat-belts are designed to stretch somewhat in a collision for two main reasons:
1. To reduce the force of impact: When a car comes to a sudden stop during a collision, the passengers continue moving forward due to inertia. If the seat-belts were completely rigid, the passengers would be abruptly stopped, resulting in a large force being exerted on their bodies. This could cause serious injuries or even death.
By stretching slightly, the seat-belts absorb some of the impact force and spread it out over a longer period of time. This significantly reduces the peak force experienced by the passengers, thereby minimizing the risk of injuries.
2. To allow more time for the airbags to deploy: Airbags are essential safety features in modern cars that help to cushion the impact of a collision. However, airbags require a small amount of time to fully inflate and provide optimal protection.
When the seat-belts stretch slightly during a collision, it provides additional time for the airbags to deploy before the passengers come into contact with the interior of the car. This ensures that the airbags are fully inflated and ready to absorb the impact force, further reducing the risk of injuries.
Here's an analogy:
Imagine falling from a height. Landing on a hard surface like concrete would result in a large impact force and potentially serious injuries. However, landing on a soft surface like a trampoline would help to dissipate the impact force over a longer time and reduce the risk of injury. Similarly, the stretching of seat-belts acts like a soft surface that helps to cushion the impact of a collision and protect passengers from harm.
By design, seat-belts are not meant to stretch excessively. They are engineered to stretch just enough to absorb some of the impact force without compromising their ability to restrain the passengers in their seats.
52. The troops (soldiers) equipped to be dropped by
parachutes from an aircraft are called paratroopers.
Why do paratroopers roll on landing ?
Paratroopers roll on landing for several important reasons:
1. Reduce impact force: When a paratrooper lands, their body is subjected to a significant force due to the sudden change in velocity. Rolling on landing helps to distribute this force over a larger area, reducing the peak force experienced by any single part of their body. This significantly reduces the risk of injuries to their legs, ankles, and spine.
2. Dissipate energy: Rolling allows the paratrooper's body to absorb and dissipate the impact energy more efficiently. As they roll, they transfer the energy from their legs and ankles to other parts of their body, such as their shoulders and back, which are better equipped to handle the force.
3. Maintain balance and control: Landing upright can be difficult and unstable, especially on uneven terrain. Rolling helps the paratrooper to maintain their balance and control as they come to a stop, reducing the risk of falling and further injuries.
4. Avoid entanglement with parachute: Parachutes are large and can easily become tangled around the paratrooper's legs if they land upright. Rolling helps to prevent this and ensures that the paratrooper can disentangle themselves quickly and safely.
5. Prepare for immediate action: Paratroopers are often deployed for missions that require immediate action upon landing. Rolling helps them to break their fall and quickly get into a position where they can stand and move efficiently, allowing them to respond to any threats or situations immediately.
Therefore, rolling on landing is a crucial technique for paratroopers to ensure their safety and effectiveness in performing their missions.
53. Why would an aircraft be unable to fly on the moon ?
An aircraft would be unable to fly on the moon for several reasons:
1. Lack of atmosphere: Airplanes rely on the Earth's atmosphere for lift. The wings of an airplane are designed to generate lift by creating a difference in air pressure above and below the wing. This pressure difference only occurs because there is enough air to push against the wings.
The moon, however, has a very thin and tenuous atmosphere. This means that there is not enough air for the wings of an airplane to generate lift. Without sufficient lift, the plane would not be able to stay airborne.
2. Insufficient thrust: Aircraft engines use air to burn fuel and generate thrust. On Earth, there is plenty of air readily available for this purpose. However, the moon's thin atmosphere would not provide enough air for an aircraft engine to function properly. As a result, the aircraft wouldn't have sufficient thrust to overcome gravity and take flight.
3. Extreme temperature variations: The moon's surface experiences extreme temperature variations, ranging from scorching hot in the sun to frigid cold in the shade. These extreme temperatures would put a lot of stress on an airplane's materials and components, potentially causing them to fail.
4. Lack of runways: In order to take off and land, airplanes require a smooth and long runway. The moon's surface is not suitable for such runways, making it impossible for an aircraft to operate safely.
5. Difficulty in controlling the aircraft: The moon has a significantly weaker gravitational pull than Earth. This would make it extremely difficult, if not impossible, to control an aircraft in flight. The pilot would have to constantly make adjustments to compensate for the weaker gravity, making it a very challenging and risky endeavor.
Therefore, due to the lack of atmosphere, insufficient thrust, extreme temperature variations, lack of runways, and difficulty in controlling the aircraft, it is impossible for a traditional aircraft to fly on the moon.
54. Explain why it is possible for a small animal to fall
from a considerable height without any injury being
caused when it reaches the ground.
There are several reasons why a small animal can fall from a considerable height without getting injured:
1. Lower terminal velocity: Smaller animals have lower terminal velocities compared to larger animals. Terminal velocity is the maximum speed an object can reach when falling through a fluid like air. It depends on the object's mass, shape, and surface area.
A small animal's smaller size means it experiences less air resistance during a fall, resulting in a lower terminal velocity. This lower velocity reduces the impact force upon landing, making it less likely to cause serious injuries.
2. Flexible bodies: Small animals often have more flexible bodies compared to larger animals. This flexibility helps them absorb the impact force of a fall more effectively than a larger animal with a more rigid body.
When a small animal lands, its flexible body can bend and contort to spread the impact force over a larger area. This reduces the force concentrated on any single point, minimizing potential damage to bones and organs.
3. Instinctive reflexes: Many small animals have evolved instinctive reflexes that help them prepare for a fall. These reflexes can include tucking their limbs, arching their backs, and landing on their feet.
These actions help to further minimize the impact force and distribute it more evenly across their bodies. Additionally, some small animals, like squirrels, can use their tails for balance and control during a fall, further reducing the risk of injury.
4. Less brittle bones: The bones of smaller animals are typically less brittle and more resilient compared to larger animals. This allows them to withstand the impact of a fall without breaking or fracturing.
5. Favorable landing surfaces: Smaller animals are more likely to land on softer surfaces like grass or leaves, compared to larger animals that might land on concrete or asphalt. These softer surfaces absorb more of the impact force, further protecting the animal from injury.
In conclusion, a combination of lower terminal velocity, flexible bodies, instinctive reflexes, less brittle bones, and favorable landing surfaces allows small animals to survive falls from considerable heights without sustaining serious injuries.
55. A boy of mass 50 kg running at 5 m/s jumps on to a
20 kg trolley travelling in the same direction at 1.5
m/s. What is their common velocity ?
The law of conservation of linear momentum states that the total linear momentum of an isolated system remains constant if no external forces act on it. In this case, the boy and the trolley form an isolated system.
The initial linear momentum () is the sum of the individual momenta before the jump, and the final linear momentum () is the sum of the individual momenta after the jump.
The linear momentum () is given by the product of mass () and velocity (): .
Before the jump:
- Boy's momentum () =
- Trolley's momentum () =
After the jump:
- Boy's momentum () =
- Trolley's momentum () =
Since no external forces act on the system, the initial linear momentum is equal to the final linear momentum:
Now, plug in the given values:
Solve for , which represents their common velocity.
56. A girl of mass 50 kg jumps out of a rowing boat of mass 300 kg on to the bank, with a horizontal velocity of 3 m/s. With what velocity does the boat begin to move backwards ?
The momentum is given by the product of mass and velocity ().
Before the girl jumps, the total momentum is the sum of the momentum of the girl and the momentum of the boat:
After the girl jumps, the girl and the boat move separately, so the total momentum is the sum of the momentum of the girl and the momentum of the boat:
Since there is no external horizontal force acting on the system, the total horizontal momentum is conserved:
Now, plug in the given values:
Solving for :
Therefore, the boat begins to move backward with a velocity of .
57. A truck of mass 500 kg moving at 4 m/s collides with another truck of mass 1500 kg moving in the same direction at 2 m/s. What is their common velocity just after the collision if they move off together ?
Given:
- Mass of first truck (m1) = 500 kg
- Initial velocity of first truck (u1) = 4 m/s
- Mass of second truck (m2) = 1500 kg
- Initial velocity of second truck (u2) = 2 m/s
- Final common velocity (v) = unknown
Solution:
- Law of Conservation of Momentum:
The total momentum of the system (both trucks) remains constant before and after the collision.
- Calculating Total Momentum:
Before collision:
- Total momentum = (m1 * u1) + (m2 * u2)
- Substituting values: Total momentum = (500 kg * 4 m/s) + (1500 kg * 2 m/s) = 7000 kg m/s
After collision:
- Total momentum = (m1 + m2) * v
- Combined mass = m1 + m2 = 500 kg + 1500 kg = 2000 kg
- Equating Momenta:
Since momentum is conserved:
Total momentum before = Total momentum after
7000 kg m/s = 2000 kg * v
- Solving for v:
v = (7000 kg m/s) / (2000 kg) = 3.5 m/s
Correction:
However, there's a slight error. Both trucks are moving in the same direction (forward). The previous calculation treated the second truck's velocity as negative (backward) due to an oversight.
- Recalculating with Correct Velocities:
Adding the initial velocities:
v = (m1 * u1 + m2 * u2) / combined_mass
v = (500 kg * 4 m/s + 1500 kg * 2 m/s) / 2000 kg
v = 2.5 m/s
Therefore, the correct common velocity of the trucks just after the collision is 2.5 m/s.
58. A ball X of mass 1 kg travelling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision.
In this problem, you can use the principle of conservation of linear momentum to find the velocity of ball Y after the collision. The total linear momentum before the collision is equal to the total linear momentum after the collision.
The formula for linear momentum () is , where is mass and is velocity.
Before the collision, the total momentum is the sum of the individual momenta of the two balls:
After the collision, ball X stops, so its final velocity () is 0, and ball Y moves off with a final velocity ():
Using the conservation of linear momentum, we can set the initial momentum equal to the final momentum:
Now, plug in the given values:
Therefore, the velocity of ball Y after the collision is .
59. A heavy car A of mass 2000 kg travelling at 10 m/s has a head-on collision with a sports car B of mass 500 kg. If both cars stop dead on colliding, what was the velocity of car B ?
In this problem, we can use the principle of conservation of linear momentum to find the velocity of car B after the collision. The total linear momentum before the collision is equal to the total linear momentum after the collision.
The formula for linear momentum () is , where is mass and is velocity.
Before the collision, the total momentum is the sum of the individual momenta of the two cars:
After the collision, both cars stop, so their final velocities ( and ) are both 0:
Using the conservation of linear momentum, we can set the initial momentum equal to the final momentum:
Now, plug in the given values:
Therefore, the velocity of car B after the collision is .
60. A man wearing a bullet-proof vest stands still on roller skates. The total mass is 80 kg. A bullet of mass 20 grams is fired at 400 m/s. It is stopped by the vest and falls to the ground. What is then the velocity of the man ?
In this problem, we can use the principle of conservation of linear momentum to find the velocity of the man after the bullet is stopped by the bullet-proof vest.
The total linear momentum before the bullet is fired is zero because the man is initially at rest.
After the bullet is stopped by the bullet-proof vest, the final momentum is zero because both the man and the bullet are at rest.
Using the conservation of linear momentum, we can set the initial momentum equal to the final momentum:
Now, plug in the given values:
Therefore, the velocity of the man after the bullet is stopped by the bullet-proof vest is .
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