Previous year questions: Laws of Motion

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Draw the free body diagram before reading the answer, and write the sign convention at the top of every solution.

Q1 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

Three masses m1 = 4 kg, m2 = 4 kg and m3 = 6 kg are suspended from a fixed smooth frictionless pulley. Masses m1 and m2 hang from the two sides of the pulley, and m3 hangs below m2. The value of T1 over T2 is (take g = 10 m s^-2)

Answer: The two sides carry 4 kg and 10 kg, so the acceleration is (10 - 4)g over 14, which is 30/7 m per second squared, with the heavier side going down. For m1 the tension T1 must support the weight and provide the acceleration: T1 equals 4 times (10 + 30/7), which is 400/7 N. For m3 the tension T2 acts upward while gravity acts down and the mass accelerates down: T2 equals 6 times (10 - 30/7), which is 240/7 N. The ratio T1/T2 is 400/240, that is 5/3.

Hinglish: Dono side 4 kg aur 10 kg hain, isliye acceleration (10-4)g/14 = 30/7 m/s^2. m1 ke liye T1 = 4(10 + 30/7) = 400/7 N. m3 ke liye T2 = 6(10 - 30/7) = 240/7 N. Ratio 400/240 = 5/3.

Q2 (JEE Main 2026, 5 Apr Shift 1, 4 marks)

A wedge Y of mass 10 kg rests on frictionless surfaces, its inclined face making 37 degrees with the horizontal. A block X of mass 2 kg is placed at the highest point of the wedge and is at rest at t = 0. The wedge is then pulled to the right by a constant horizontal force of 24 N. The time taken by the block to slide 8.8 m down the slope is (take tan 37 = 3/4 and g = 10 m s^-2)

Answer: The acceleration of the wedge is 24/10, that is 2.4 m per second squared to the right. In the frame of the wedge the block carries a pseudo force of mass times wedge acceleration to the left, with magnitude 2 times 2.4, that is 4.8 N. Along the incline the weight component is mg sin 37, which is 2 times 10 times 0.6, equal to 12 N down the slope; the pseudo force component along the slope is 4.8 cos 37, about 3.84 N up the slope. The net downslope force is about 8.16 N, so the acceleration along the slope is 4.08 m per second squared. From s = half a t squared, 8.8 equals 2.04 t squared, so t is about 2 s.

Hinglish: Wedge ka acceleration 24/10 = 2.4 m/s^2. Wedge frame mein block par pseudo force 2 x 2.4 = 4.8 N left. Slope ke along weight component 2 x 10 x 0.6 = 12 N neeche, aur pseudo ka component 4.8 x 0.8 = 3.84 N upar. Net 8.16 N, acceleration 4.08 m/s^2. 8.8 = 2.04 t^2 se t ≈ 2 s.

Q3 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A cube moves to the right with acceleration g/2. Blocks P of mass 2 kg and Q of mass 1.5 kg are joined by a massless thread over a frictionless pulley fixed at the edge of the cube. P sits on the frictionless top surface and Q touches the rough side face without moving relative to the cube. The coefficient of friction between Q and the side face is (take g = 10 m s^-2)

Answer: In the frame of the cube the pseudo force on P is 2 times g/2, that is 10 N to the left, and since P stays at rest relative to the cube the tension must be 10 N. For Q the pseudo force presses it against the side face with normal reaction 1.5 times g/2, that is 7.5 N. Q's weight is 15 N downward while the tension lifts with 10 N, leaving 5 N that friction must balance. So the friction coefficient is 5 over 7.5, about 0.67.

Hinglish: Cube frame mein P par pseudo force 2 x g/2 = 10 N, isliye tension 10 N. Q ke liye normal 1.5 x g/2 = 7.5 N; weight 15 N neeche, tension 10 N upar, baaki 5 N friction balance karta hai. mu = 5/7.5 ≈ 0.67.

Q4 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A block takes time t to slide down a plane inclined at 45 degrees to the horizontal. If the surface is made smooth, the block takes time t/2 to slide down the same plane. The coefficient of friction is a/100, where a is

Answer: For the same distance s, the relation s = half a t squared gives a times t squared constant. The smooth acceleration is g sin 45, and the rough acceleration is g(sin 45 minus mu cos 45), which is g(1 - mu) over root 2. Setting the products equal: g(1 - mu)t squared over root 2 equals g t squared over 4 root 2. The factor cancels and 1 - mu = 1/4, so mu = 3/4 = 0.75 and a = 75.

Hinglish: Same distance ke liye a t^2 constant hai. Smooth acceleration g/root2, rough acceleration g(1-mu)/root2. Dono ke product barabar rakho: g(1-mu)t^2/root2 = g t^2/(4 root2), isliye 1 - mu = 1/4, mu = 0.75, matlab a = 75.

Q5 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

The time taken by a block to slide from the highest point to the lowest point of a rough inclined plane at 45 degrees is 50 percent more than the time taken on an identical smooth plane. The coefficient of kinetic friction is

Answer: The rough time is 1.5 times the smooth time, so for the same distance the accelerations satisfy a_rough times (1.5) squared equal a_smooth. That gives a_rough = 4a_smooth/9. The smooth acceleration is g sin 45 and the rough one is g(1 - mu) over root 2, so 1 - mu = 4/9 and mu = 5/9.

Hinglish: Rough time smooth ka 1.5 guna hai, isliye same distance par a_rough (1.5)^2 = a_smooth, matlab a_rough = 4a_smooth/9. Smooth acceleration g/root2, rough g(1-mu)/root2, isliye 1 - mu = 4/9 aur mu = 5/9.

Q6 (JEE Main 2026, 8 Apr Shift 2, 4 marks)

A car moving at 54 km per hour takes a turn of radius 20 m. A simple pendulum hangs from the ceiling of the car. The angle made by the string with the vertical during the turn is (take g = 10 m s^-2)

Answer: Convert the speed: 54 km per hour is 15 m per second. In the car's frame the bob is in equilibrium under its weight, the tension and the pseudo force outward, which is mass times centripetal acceleration, m v squared over r. The angle with the vertical satisfies tan theta equal to v squared over r g, which is 225 over 200, that is 1.125. So the angle is the inverse tangent of 1.125, and the option carrying tan inverse 1.125 is correct.

Hinglish: Speed 54 km/h = 15 m/s. Car frame mein bob par weight, tension aur pseudo force mv^2/r outward lagta hai. tan theta = v^2/(rg) = 225/200 = 1.125, isliye angle tan^-1(1.125) hai.

Q7 (NCERT Exemplar, 1 mark)

A ball is travelling with uniform translatory motion. This means that (a) it is at rest (b) the path can be a straight line or circular and the ball travels with uniform speed (c) all parts of the ball have the same velocity, magnitude and direction, and the velocity is constant (d) the centre of the ball moves with constant velocity and the ball spins about its centre uniformly

Answer: Translatory motion with no rotation is what uniform translatory motion means, so every part of the ball has the same velocity at every instant. Option (b) allows a circular path, which is not translatory and also changes direction, so it cannot be uniform translatory motion. Option (d) adds a spin, which is rotation and not translation. Option (c) is therefore correct.

Hinglish: Uniform translatory motion mein body ke saare parts ki velocity same hoti hai, koi spin nahi. Option (b) circular path allow karta hai jo translatory nahi, aur (d) spin add karta hai jo rotation hai. Isliye (c) correct hai.

Q8 (NCERT Exemplar, 1 mark)

Conservation of momentum in a collision between particles can be understood from (a) conservation of energy (b) Newton's first law only (c) Newton's second law only (d) both Newton's second and third law

Answer: During the collision the third law makes the internal forces equal and opposite, so their impulses cancel in the total. The second law then says that with no net external force the total momentum does not change. Both laws are needed, so option (d) is correct. Energy conservation is a separate statement and does not by itself imply momentum conservation.

Hinglish: Collision ke andar third law se internal forces equal opposite hote hain, isliye unke impulses cancel ho jaate hain. Second law se net external force zero hone par total momentum nahi badalta. Dono laws chahiye, isliye (d) sahi hai.

Q9 (NCERT Exemplar, 1 mark)

A body of mass 2 kg travels according to x(t) = pt + qt squared + rt cubed, with p = 1 m s^-1, q = 4 m s^-2 and r = 5 m s^-3. The force acting on the body at t = 2 s is

Answer: Differentiate twice. The velocity is p + 2qt + 3rt squared and the acceleration is 2q + 6rt. At t = 2 s the acceleration is 8 + 60, that is 68 m per second squared. The force is mass times acceleration, 2 times 68, which is 136 N. The option 136 N is correct.

Hinglish: Do baar differentiate karo: velocity p + 2qt + 3rt^2, acceleration 2q + 6rt. t = 2 par 8 + 60 = 68 m/s^2. Force 2 x 68 = 136 N.

Q10 (NCERT Exemplar, 1 mark)

A car of mass m starts from rest and acquires a velocity v along the east in 2 s with uniform acceleration. The force exerted on the car is (a) mv/2 eastward, exerted by the engine (b) mv/2 eastward, due to friction on the tyres exerted by the road (c) more than mv/2 eastward, due to the engine overcoming friction (d) mv, exerted by the engine

Answer: The acceleration is v/2, so the net force on the car is mv/2 eastward. The engine turns the wheels, and the road exerts a forward friction force on the tyres; it is this external friction that accelerates the car, since internal engine forces cannot change the momentum of the car as a whole. Option (b) is correct. The engine force acts on the wheels, but the force on the car from outside is the friction from the road.

Hinglish: Acceleration v/2, isliye net force mv/2 eastward. Engine wheels ko ghumata hai, aur road tyres par forward friction lagata hai; yahi external friction car ko accelerate karta hai. Internal engine force car ke total momentum ko nahi badal sakta. Isliye (b) sahi hai.