CBSE Class 10 Physics Force and Laws of Motion MCQs Set 22

Physics Objective Questions and Answers: Force and Laws of Motion

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Multiple Choice Questions

Question 1: A body of mass m collides against a wall with a velocity v and retraces its path with the same speed. The magnitude of the change in momentum is
(a) zero
(b) 2 mv
(c) mv
(d) \( -mv \)
Show Answer & Explanation

Answer: (b) 2 mv

Explanation:
1. Initial momentum = +mv, final momentum = -mv.
2. Change = -mv - mv = -2mv, so the magnitude is 2mv.

Teacher's Note:
1. Reversal of direction changes the sign of the momentum.
2. The change is twice the initial momentum.

Question 2: In SI units a force is numerically equal to the ____, when the force is applied to it.
(a) velocity of one kilogram object
(b) speed of one kilogram object
(c) acceleration of one kilogram object
(d) acceleration of any object
Show Answer & Explanation

Answer: (c) acceleration of one kilogram object

Explanation:
1. One newton gives 1 kg an acceleration of 1 m/s\( ^2 \).
2. So the numerical value of F equals the acceleration of a 1 kg object.

Teacher's Note:
1. F = ma with m = 1 kg gives F = a numerically.

Question 3: A body of mass 2 kg moving with a speed of 100 m/s hits a wall and rebounds with the same speed. If the contact time is \( \tfrac{1}{50} \) s, the force applied on the wall is
(a) \( 10^4 \) N
(b) \( 2\times10^4 \) N
(c) 0 N
(d) \( 4\times10^4 \) N
Show Answer & Explanation

Answer: (b) \( 2\times10^4 \) N

Explanation:
1. Change in momentum = 2 x 100 - (2 x -100) = 400 kg m/s.
2. F = 400 / (1/50) = 20000 N = \( 2\times10^4 \) N.

Teacher's Note:
1. The rebound reverses the velocity, so the change is twice mv.

Question 4: A cricket ball of mass 150 g moves at a speed of 12 m/s and, after hitting by the bat, is deflected back at a speed of 20 m/s. If the bat and the ball remained in contact for 0.02 s, the impulse and average force exerted by the bat on the ball are respectively
(a) 2.4 N s ; 480 N
(b) 2.4 N s ; 120 N
(c) 4.8 N s ; 240 N
(d) 4.8 N s ; 480 N
Show Answer & Explanation

Answer: (c) 4.8 N s ; 240 N

Explanation:
1. Mass = 0.15 kg. Take the final direction as positive: change in velocity = 20 - (-12) = 32 m/s.
2. Impulse = 0.15 x 32 = 4.8 N s.
3. Force = 4.8 / 0.02 = 240 N.

Teacher's Note:
1. The velocity reverses, so the two speeds add.
2. Convert grams to kilograms first.

Question 5: The velocity-time graph of a ball of mass 50 g rolling on a concrete floor is shown in the figure: the velocity falls uniformly from 80 m/s at t = 0 to 0 at t = 8 s. The ball comes to rest due to the frictional force of the floor on it. This frictional force is
Time (s)Velocity (m/s)2040608002468 (a) 1.5 N
(b) 0.5 N
(c) 1 N
(d) 2 N
Show Answer & Explanation

Answer: (b) 0.5 N

Explanation:
1. Slope of the graph = (0 - 80)/8 = -10 m/s\( ^2 \).
2. Retardation = 10 m/s\( ^2 \).
3. F = ma = 0.05 x 10 = 0.5 N.

Teacher's Note:
1. Slope of a velocity-time graph gives acceleration.
2. Convert 50 g to 0.05 kg.

Question 6: A force of 5 N produces an acceleration of 8 m s\( ^{-2} \) on a mass \( m_1 \) and an acceleration of 24 m s\( ^{-2} \) on a mass \( m_2 \). What acceleration (in m s\( ^{-2} \)) would the same force provide if both the masses are tied together?
(a) 12
(b) 9
(c) 10
(d) 6
Show Answer & Explanation

Answer: (d) 6

Explanation:
1. \( m_1 = 5/8 \) kg and \( m_2 = 5/24 \) kg.
2. Total mass = 5/8 + 5/24 = 20/24 = 5/6 kg.
3. a = 5 / (5/6) = 6 m/s\( ^2 \).

Teacher's Note:
1. Add the masses, not the accelerations.
2. Heavier combined mass gives a smaller acceleration.

Question 7: A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object's velocity from 3 m s\( ^{-1} \) to 7 m s\( ^{-1} \). If the force were applied for a duration of 5 s, what would be the final velocity of the object?
(a) 11 m/s
(b) 12 m/s
(c) 13 m/s
(d) 14 m/s
Show Answer & Explanation

Answer: (c) 13 m/s

Explanation:
1. Acceleration = (7 - 3)/2 = 2 m/s\( ^2 \).
2. For 5 s: v = u + at = 3 + 2 x 5 = 13 m/s.

Teacher's Note:
1. The acceleration is the same because the force and mass are the same.

Question 8: A bullet of mass 0.04 kg moving with a speed of 90 m s\( ^{-1} \) enters a heavy wooden block and is stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet?
(a) 230 N
(b) 260 N
(c) 540 N
(d) 270 N
Show Answer & Explanation

Answer: (d) 270 N

Explanation:
1. Use \( v^2 = u^2 + 2as \): \( 0 = 90^2 + 2a(0.6) \), so \( a = -6750 \) m/s\( ^2 \).
2. F = ma = 0.04 x 6750 = 270 N.

Teacher's Note:
1. Change 60 cm to 0.6 m.
2. The negative sign only shows that the force opposes the motion.

Question 9: A force of magnitude F acts on a body of mass m. The acceleration of the body depends upon
(a) volume of body
(b) density of body
(c) area of body
(d) mass of body
Show Answer & Explanation

Answer: (d) mass of body

Explanation:
1. By the second law \( a = F/m \).
2. For a given force only the mass matters.

Teacher's Note:
1. Volume, density and area do not appear in the formula.

Question 10: Force remaining constant, if the mass of a body increases, its acceleration is likely to
(a) increase
(b) remain the same
(c) decrease
(d) sometimes increase and sometimes decrease
Show Answer & Explanation

Answer: (c) decrease

Explanation:
1. \( a = F/m \).
2. Larger m with the same F gives smaller a.

Teacher's Note:
1. Acceleration is inversely proportional to mass for a given force.

Question 11: A force F produces an acceleration a. The same force produces an acceleration 4a in another body. The mass of the other body is
(a) four times the mass of the first body
(b) four times smaller than the mass of the first body
(c) mass does not play any role
(d) none of the above
Show Answer & Explanation

Answer: (b) four times smaller than the mass of the first body

Explanation:
1. \( m = F/a \). For the second body \( m' = F/(4a) = m/4 \).

Teacher's Note:
1. Four times the acceleration needs one quarter of the mass.

Question 12: The mass of an aeroplane is 2.5 tonnes. Its engine develops a force of 8750 N before taking off. The acceleration of the aeroplane at the time of take off is
(a) 3.45 m s\( ^{-2} \)
(b) 3.65 m s\( ^{-2} \)
(c) 3.50 m s\( ^{-2} \)
(d) 3.60 m s\( ^{-2} \)
Show Answer & Explanation

Answer: (c) 3.50 m s\( ^{-2} \)

Explanation:
1. Mass = 2.5 tonnes = 2500 kg.
2. a = F/m = 8750 / 2500 = 3.5 m/s\( ^2 \).

Teacher's Note:
1. 1 tonne = 1000 kg.

Question 13: When a force of 1 N acts on a mass of 1 kg that is able to move freely, the object moves with
(a) a speed of 1 m/s
(b) a speed of 1 km/s
(c) an acceleration of 10 m/s\( ^2 \)
(d) an acceleration of 1 m/s\( ^2 \)
Show Answer & Explanation

Answer: (d) an acceleration of 1 m/s\( ^2 \)

Explanation:
1. This is the definition of 1 newton.
2. a = F/m = 1/1 = 1 m/s\( ^2 \).

Teacher's Note:
1. A force produces acceleration, not a fixed speed.

Question 14: The rate of change of momentum gives
(a) velocity
(b) impulse
(c) force
(d) acceleration
Show Answer & Explanation

Answer: (c) force

Explanation:
1. Newton's second law: \( F = \Delta p/\Delta t \).

Teacher's Note:
1. Impulse is the change in momentum itself, not its rate.

Question 15: China wares are wrapped in straw or paper before packing. This
(a) increases the time of impact
(b) decreases the change in momentum
(c) increases the acceleration
(d) increases the force
Show Answer & Explanation

Answer: (a) increases the time of impact

Explanation:
1. In a jolt the change in momentum is fixed.
2. The soft wrapping lengthens the time of impact, so the force is smaller (\( F = \Delta p/t \)).

Teacher's Note:
1. Soft padding spreads the stopping over a longer time.

Question 16: A force of 20 N acting on a body of mass 10 kg is found to double its velocity in 8 s. Find its initial velocity.
(a) 6 m s\( ^{-1} \)
(b) 10 m s\( ^{-1} \)
(c) 16 m s\( ^{-1} \)
(d) 20 m s\( ^{-1} \)
Show Answer & Explanation

Answer: (c) 16 m s\( ^{-1} \)

Explanation:
1. a = F/m = 20/10 = 2 m/s\( ^2 \).
2. Final velocity = 2u, so \( 2u = u + 2\times8 \).
3. u = 16 m/s.

Teacher's Note:
1. Use \( v = u + at \) with v = 2u.

Question 17: The unit of impulse is
(a) newton
(b) newton second
(c) newton second\( ^2 \)
(d) newton per second
Show Answer & Explanation

Answer: (b) newton second

Explanation:
1. Impulse = force x time, so the unit is N s (same as kg m/s).

Teacher's Note:
1. Impulse and momentum have the same units.

Question 18: The variation of force with time for a body moving along a straight line is shown in the figure: the force rises uniformly from -2 N at t = 0 to +2 N at t = 4 s and then falls uniformly back to -2 N at t = 8 s. The impulse (in N s) imparted in the time interval t = 0 to t = 8 s is
t (s)F (N)+2-22468 (a) zero
(b) 8
(c) -8
(d) 6
Show Answer & Explanation

Answer: (a) zero

Explanation:
1. Impulse = area under the force-time graph, counted with sign.
2. From 0 to 2 s: area = -2 N s. From 2 to 6 s: +4 N s. From 6 to 8 s: -2 N s.
3. Total = 0.

Teacher's Note:
1. The graph is symmetric about t = 4 s with positive and negative areas that cancel.

Question 19: A revolver recoils after a bullet is fired because
(a) the man draws it backwards out of fear
(b) the trigger of the revolver moves backwards
(c) the forward motion of the bullet imparts to the revolver an equal and opposite reaction
(d) the question is irrelevant
Show Answer & Explanation

Answer: (c) the forward motion of the bullet imparts to the revolver an equal and opposite reaction

Explanation:
1. The revolver pushes the bullet forward.
2. By the third law the bullet pushes the revolver backward with an equal force.

Teacher's Note:
1. Recoil is also explained by conservation of momentum.

Question 20: A porter is carrying a weight of 200 N on his head. If the force exerted on his head by the weight is taken as action, then the reaction force is exerted by
(a) the head on the weight
(b) the weight on the earth
(c) the earth on the porter
(d) the earth on the weight
Show Answer & Explanation

Answer: (a) the head on the weight

Explanation:
1. Action: the weight pushes down on the head.
2. Reaction: the head pushes up on the weight, equal and opposite.

Teacher's Note:
1. An action-reaction pair acts between the same two bodies, with roles reversed.

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