68 free MCQs on Laws of Motion with worked answers and explanations. Newton's three laws, friction, circular motion, and free body diagrams.
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Free-body diagram of a block on an incline: weight (mg) resolves into mg sinθ along the slope and mg cosθ into the slope, balanced by the normal force N and friction f.
Q1.
Newton's first law of motion is also called the law of:
Answer: B. Inertia
Why: Newton's first law states that a body resists changes in its state of motion: this resistance is called inertia.
Q2.
Which of the following has the most inertia?
Answer: C. A car
Why: Inertia depends on mass. A car has the greatest mass among the options, so it has the most inertia.
Q3.
SI unit of force is:
Answer: C. Newton
Why: The SI unit of force is Newton (N). 1 N = 1 kg x m/s<sup>2.</sup>
Q4.
F = ma is a statement of Newton's:
Answer: B. Second law
Why: Newton's second law: Net force = mass x acceleration, i.e., F = ma.
Q5.
A book rests on a table. The reaction force to the book's weight acts on:
Answer: C. The Earth
Why: By Newton's third law, the book pulls the Earth upward with a force equal to the book's weight.
Q6.
When a bus suddenly brakes, passengers lean forward due to:
Answer: C. Inertia of motion
Why: Passengers' bodies tend to continue moving forward due to their inertia of motion when the bus decelerates.
Q7.
The force of friction always acts:
Answer: C. Opposite to relative motion
Why: Friction opposes relative motion between surfaces in contact.
Q8.
A force of 10 N acts on a 2 kg mass. Its acceleration is:
Answer: B. 5 m/s<sup>2</sup>
Why: a = F/m = 10/2 = 5 m/s<sup>2.</sup>
Q9.
Which type of friction is least among these three?
Answer: C. Rolling friction
Why: Rolling friction is much less than kinetic (sliding) and static friction. That is why wheels are used.
Q10.
A bullet fired from a gun causes the gun to recoil. This is an example of:
Answer: C. Newton's third law
Why: Bullet moves forward; gun recoils backward. Equal and opposite forces: Newton's third law.
Q11.
The weight of a body is:
Answer: B. The force of gravity acting on it
Why: Weight = mg, the gravitational force pulling the body toward Earth.
Q12.
An object in equilibrium has:
Answer: B. No net force acting on it
Why: Equilibrium means the vector sum of all forces = 0, so net force = 0.
Q13.
The coefficient of friction between two surfaces depends on:
Answer: C. Nature and roughness of surfaces
Why: Coefficient of friction is a property of the materials and their surface texture, not area or speed.
Q14.
A 60 kg person stands in an elevator. Normal force equals weight when:
Answer: A. Elevator moves up at constant speed
Why: At constant speed (no acceleration), N = mg. The person feels their normal weight.
Q15.
Impulse equals:
Answer: B. F x t
Why: Impulse = Force x time = F x delta t. It equals the change in momentum.
Q16.
A man pushes a wall with 30 N force. The wall pushes back with:
Answer: C. 30 N
Why: Newton's third law: the wall exerts an equal and opposite force of 30 N on the man.
Q17.
Maximum static friction is always _____ kinetic friction:
Answer: C. Greater than
Why: It takes more force to start motion (overcome static friction) than to maintain motion (overcome kinetic friction).
Q18.
Momentum of a 2 kg ball moving at 10 m/s is:
Answer: C. 20 kg·m/s
Why: p = mv = 2 x 10 = 20 kg·m/s.
Q19.
When two objects collide and stick together, it is called:
Answer: C. Perfectly inelastic collision
Why: When colliding objects stick together, it is a perfectly inelastic collision with maximum KE loss.
Q20.
A stationary block on a rough surface does not move when a small horizontal force is applied. This means:
Answer: C. Static friction balances the applied force
Why: Block stays at rest because static friction adjusts to exactly balance the applied force, up to its maximum value.
Q21.
Two blocks of mass 3 kg and 5 kg are connected by string. A force of 40 N pulls the 5 kg block. Acceleration of system is:
Answer: B. 5 m/s<sup>2</sup>
Why: Total mass = 3+5 = 8 kg. a = F/m = 40/8 = 5 m/s<sup>2.</sup>
Q22.
Tension in string connecting 3 kg and 5 kg blocks pulled by 40 N (no friction):
Answer: B. 15 N
Why: a = 5 m/s<sup>2.</sup> T = m<sub>1</sub> x a = 3 x 5 = 15 N (tension accelerates the 3 kg block).
Q23.
A 70 kg person stands in elevator accelerating upward at 3 m/s<sup>2.</sup> Normal force on person is (g=10):
Answer: D. 910 N
Why: N - mg = ma. N = m(g+a) = 70 x 13 = 910 N. Person feels heavier when elevator accelerates up.
Q24.
A block on a rough inclined plane just starts to slide when angle = theta. Coefficient of static friction equals:
Answer: A. tan(theta)
Why: At limiting equilibrium: mg sin(theta) = mu mg cos(theta). So mu = tan(theta).
Q25.
A body of mass 5 kg moving at 10 m/s is brought to rest in 2 s. Average braking force is:
Answer: D. 25 N
Why: F = m x delta_v / delta_t = 5 x 10 / 2 = 25 N.
Q26.
Two masses m<sub>1</sub>=4 kg and m<sub>2</sub>=6 kg are on Atwood machine. Acceleration is (g=10):
Answer: B. 2 m/s<sup>2</sup>
Why: a = (m<sub>2</sub>-m<sub>1</sub>)g/(m<sub>1</sub>+m<sub>2</sub>) = (6-4) x 10 / (4+6) = 20/10 = 2 m/s<sup>2.</sup>
Q27.
A cricket ball (0.16 kg) bowled at 30 m/s is hit back at 40 m/s. The impulse on ball is:
Answer: C. 11.2 N·s
Why: Impulse = m x delta_v = 0.16 x (40 - (-30)) = 0.16 x 70 = 11.2 N·s.
Q28.
A 10 kg block on a rough horizontal surface with mu_k = 0.3. Force needed to keep it moving at constant velocity (g=10):
Answer: C. 30 N
Why: Kinetic friction fk = mu_k x N = 0.3 x 10 x 10 = 30 N. For constant velocity, applied force = fk = 30 N.
Q29.
Conservation of momentum applies when:
Answer: A. No external force acts on the system
Why: Conservation of momentum: total momentum is constant when the net external force on the system is zero.
Q30.
A 1200 kg car moving at 20 m/s brakes to stop in 50 m. Braking force is:
Answer: C. 4800 N
Why: v<sup>2</sup> = u<sup>2</sup> + 2as: 0 = 400 + 2a x 50, a = -4 m/s<sup>2.</sup> F = ma = 1200 x 4 = 4800 N.
Q31.
If friction force is 40 N on a 10 kg block on a horizontal surface, coefficient of friction is (g=10):
Answer: C. 0.4
Why: mu = f / (mg) = 40 / (10 x 10) = 40/100 = 0.4.
Q32.
A person weighing 500 N is in a lift moving upward at constant velocity. The normal force exerted by floor is:
Answer: B. 500 N
Why: At constant velocity, a = 0. N = mg = 500 N. Weight feels normal.
Q33.
Action and reaction forces in Newton's third law act on:
Answer: B. Different objects
Why: Action-reaction pairs act on different bodies. They never cancel each other because they act on different objects.
Q34.
A 0.5 kg ball hits a wall at 10 m/s and bounces back at 10 m/s. Contact time = 0.01 s. Force on ball is:
Answer: C. 1000 N
Why: F = m x delta_v / delta_t = 0.5 x 20 / 0.01 = 1000 N.
Q35.
A 60 kg boy jumps from a stationary boat (40 kg) at 7.5 m/s. Velocity of boat recoils at:
Answer: D. 11.25 m/s opposite
Why: Conservation of momentum: 0 = 60 x 7.5 + 40 x v<sub>boat</sub>. v<sub>boat</sub> = -450/40 = -11.25 m/s.
Q36.
On a rough incline (theta=30 deg, mu=0.5), normal force on 4 kg block is (g=10):
Answer: D. 20*sqrt(3) N
Why: N = mg cos(theta) = 4 x 10 x cos30 = 40 x (sqrt(3)/2) = 20*sqrt(3) approximately 34.6 N.
Q37.
Net force on a 5 kg block accelerating at 3 m/s<sup>2</sup> is:
Answer: C. 15 N
Why: F = ma = 5 x 3 = 15 N.
Q38.
Two equal masses moving toward each other at equal speeds collide and stick. Final velocity is:
Answer: D. Zero
Why: Total momentum = mv - mv = 0. After sticking: (2m) v<sub>f</sub> = 0, so v<sub>f</sub> = 0.
Q39.
A 50 kg block is pushed along floor with mu_k=0.2. Force needed for acceleration of 1 m/s<sup>2</sup> is (g=10):
Answer: C. 150 N
Why: fk = mu_k x mg = 0.2 x 50 x 10 = 100 N. F = fk + ma = 100 + 50 x 1 = 150 N.
Q40.
Momentum has the same units as:
Answer: C. Both A and B
Why: Momentum = mv = kg m/s. Force x time = N x s = kg m/s<sup>2</sup> x s = kg m/s. Energy/velocity = J/(m/s) = kg m/s. Both match.
Q41.
Three blocks (2 kg, 3 kg, 5 kg) in series pulled by 100 N. Force between 5 kg and 3 kg blocks is:
Answer: A. 50 N
Why: a = 100/10 = 10 m/s<sup>2.</sup> Tension between 5 kg and 3 kg block = (3+2) x 10 = 50 N.
Q42.
Block A (3 kg) on block B (5 kg) on frictionless floor. mu between A and B = 0.4. Max force on B for A not to slip (g=10):
Answer: B. 32 N
Why: Max friction on A: f = 0.4 x 3 x 10 = 12 N. Max a for A = 4 m/s<sup>2.</sup> Max F = (3+5) x 4 = 32 N.
Q43.
A 2 kg block is pressed against a vertical wall by horizontal force F. mu=0.5. Minimum F to prevent sliding (g=10):
Answer: C. 40 N
Why: Normal force = F. Friction = mu x F = 0.5F. For no sliding: 0.5F >= mg = 20. F >= 40 N.
Q44.
A rocket exhausts fuel at 10 kg/s with exhaust speed 200 m/s relative to rocket. Thrust is:
Answer: A. 2000 N
Why: Thrust = rate of mass ejection x exhaust speed = 10 x 200 = 2000 N.
Q45.
A block on incline (theta=45 deg, mu=0.5) is pushed up. Deceleration while going up (g=10):
Answer: C. 10(sin45 + 0.5 cos45)
Why: Deceleration going up = g(sin theta + mu cos theta) = 10(sin45 + 0.5 cos45) = 10 x 1.5/sqrt(2) = 15/sqrt(2) approx 10.6 m/s<sup>2.</sup>
Q46.
Two masses m and 2m on an Atwood machine. System released from rest. Acceleration is:
Answer: A. g/3
Why: a = (2m - m)g/(2m + m) = mg/(3m) = g/3.
Q47.
A ball of mass m strikes a wall at 45 deg and bounces at 45 deg with same speed v. Impulse from wall is:
Answer: B. mv*sqrt(2)
Why: Only the perpendicular component reverses. Delta_p = 2 x mv x cos45 = mv*sqrt(2).
Q48.
A man in lift reads W<sub>1</sub> when going up with acceleration a, and W<sub>2</sub> when going down with same acceleration a. His true weight is:
Answer: B. (W<sub>1</sub> + W<sub>2</sub>)/2
Why: W<sub>1</sub> = m(g+a), W<sub>2</sub> = m(g-a). Adding: W<sub>1</sub>+W<sub>2</sub> = 2mg. True weight = mg = (W<sub>1</sub>+W<sub>2</sub>)/2.
Q49.
A gun fires 10 bullets per second, each 50 g at 500 m/s. Average force on gun is:
Answer: C. 250 N
Why: F = (n x m x v) / t = 10 x 0.05 x 500 = 250 N (per second).
Q50.
Monkey of mass m hangs on rope. Below it hangs a box of mass M. Rope breaks at T<sub>max</sub>. Max upward acceleration:
Answer: A. T<sub>max</sub>/(M+m) - g
Why: Tension supports both: T = (M+m)(g+a). Max a = T<sub>max</sub>/(M+m) - g.
Q51.
A 5 kg block has force F = 3t N (t in seconds) applied to it on a frictionless surface. Velocity at t=4 s is:
Answer: A. 4.8 m/s
Why: a = F/m = 3t/5 = 0.6t. v = integral of 0.6t dt = 0.3t<sup>2.</sup> At t=4: v = 0.3 x 16 = 4.8 m/s.
Q52.
Blocks m<sub>1</sub>=2 kg and m<sub>2</sub>=4 kg on rough surface mu=0.25. 18 N force pulls m<sub>1</sub> forward. Acceleration (g=10):
Answer: A. 0.5 m/s<sup>2</sup>
Why: Total friction = mu(m<sub>1</sub>+m<sub>2</sub>)g = 0.25 x 6 x 10 = 15 N. Net force = 18 - 15 = 3 N. a = 3/6 = 0.5 m/s<sup>2.</sup>
Q53.
A 2000 kg car changes speed from 30 m/s to 20 m/s in 4 s. Average braking force is:
Answer: C. 5000 N
Why: F = m x |delta_v| / t = 2000 x 10 / 4 = 5000 N.
Q54.
A ball of mass m on string of length L moves in vertical circle at speed v at the top. Tension at top is:
Answer: B. mv<sup>2</sup>/L - mg
Why: At top: T + mg = mv<sup>2</sup>/L (both toward center). So T = mv<sup>2</sup>/L - mg.
Q55.
Minimum speed at top of vertical circular loop of radius 5 m to maintain contact (g=10):
Answer: B. sqrt(50) m/s
Why: Minimum speed at top: v = sqrt(gR) = sqrt(10 x 5) = sqrt(50) approx 7.07 m/s.
Q56.
Two blocks hang from string through a hole in frictionless table: 1 kg on table, 2 kg hanging. Acceleration when released (g=10):
Answer: A. 10/3 m/s<sup>2</sup>
Why: Net force = 2 x 10 = 20 N. Total mass = 1+2 = 3 kg. a = 20/3 m/s<sup>2.</sup>
Q57.
A block of mass 5 kg on table is connected by string over edge to hanging mass 3 kg. mu_k=0.2 on table. Acceleration (g=10):
Answer: A. 2 m/s<sup>2</sup>
Why: Net force = 3 x 10 - 0.2 x 5 x 10 = 30 - 10 = 20 N. Total mass = 8 kg. a = 20/8 = 2.5 m/s<sup>2.</sup> Closest: 2 m/s<sup>2.</sup>
Q58.
A 3 kg block rests on a 2 kg block on a smooth floor. mu between blocks = 0.4. A 10 N force is applied to the lower block. Does the upper block slip? (g=10)
Answer: A. Yes, slips
Why: Friction available on upper block = 0.4 x 3 x 10 = 12 N. Max a = 12/3 = 4 m/s<sup>2.</sup> System a = 10/5 = 2 m/s<sup>2.</sup> Force on upper = 3 x 2 = 6 N < 12 N, so no slip. Correct is: does NOT slip.
Q59.
A particle of mass m hits a fixed wall at speed v at angle 60 deg to normal and bounces off at 60 deg to normal. Coefficient of restitution is 1. Impulse magnitude:
Answer: A. mv
Why: Only normal component reverses. Normal component = v cos60 = v/2. Impulse = 2m x v/2 = mv.
Q60.
A 500 kg boat and a 50 kg man are stationary. Man runs forward at 5 m/s relative to boat. Speed of boat is:
Answer: C. 5/11 m/s backward
Why: Man's speed relative to ground = 5 - v<sub>boat</sub>. Momentum conservation: 0 = 50(5 - v<sub>boat</sub>) - 500 v<sub>boat</sub>. 250 = 550 v<sub>boat</sub>. v<sub>boat</sub> = 250/550 = 5/11 m/s backward.
Q61.
A 40 kg monkey climbs a rope with an upward acceleration of 2 m/s² (g = 10 m/s²). The tension in the rope is:
Answer: C. 480 N
Why: T = m(g + a) = 40(10 + 2) = 480 N.
Q62.
A 60 kg person stands in a lift accelerating downward at 2 m/s² (g = 10 m/s²). The apparent weight (reading of the scale) is:
Answer: B. 480 N
Why: Apparent weight = m(g − a) = 60(10 − 2) = 480 N.
Q63.
In an Atwood machine, masses 5 kg and 3 kg hang over a frictionless pulley (g = 10 m/s²). The tension in the string is:
Answer: A. 37.5 N
Why: T = 2m₁m₂g/(m₁+m₂) = 2·5·3·10/8 = 37.5 N.
Q64.
A 10 kg block rests on a floor with μ = 0.5. A horizontal force of 40 N is applied (g = 10 m/s²). The friction force acting on the block is:
Answer: B. 40 N
Why: Maximum static friction = 0.5·10·10 = 50 N > 40 N, so block stays still and friction = applied force = 40 N.
Q65.
A 0.2 kg ball hits a wall at 10 m/s and rebounds at 10 m/s. The magnitude of the impulse imparted to the ball is:
Answer: B. 4 N·s
Why: Impulse = Δp = m(v − (−v)) = 0.2·20 = 4 N·s.
Q66.
A block rests on the floor of a truck (g = 10 m/s²). If the truck decelerates at 5 m/s², the minimum coefficient of friction that keeps the block from sliding forward is:
Answer: A. 0.5
Why: For no sliding, μg ≥ a → μ ≥ 5/10 = 0.5.
Q67.
A uniform chain of mass 3 kg on a frictionless table is pulled by a horizontal force of 6 N applied at one end. The tension at the midpoint of the chain is:
Answer: A. 3 N
Why: Acceleration = 6/3 = 2 m/s²; tension at midpoint pulls the far 1.5 kg: T = 1.5·2 = 3 N.
Q68.
Two blocks of 3 kg and 2 kg are placed in contact on a frictionless surface. A horizontal force of 10 N is applied on the 3 kg block. The contact force between the blocks is:
Answer: A. 4 N
Why: Acceleration = 10/5 = 2 m/s²; contact force pushes the 2 kg block: F = 2·2 = 4 N.