Laws of Motion NEET PYQReal NEET & AIPMT 2005–2021 questions, year-tagged, with solutions
By the RankUp team, founded by Dr. Deepak Sahu (MBBS Gold Medalist, NEET PG AIR 1731) · Updated
Every Laws of Motion question from the NEET & AIPMT papers on this site (2005–2021), newest first. Each is tagged with its paper. Tap an option to check your answer and read the explanation.
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Laws of Motion previous year questions
30 questions. Tap an option to check your answer.
Q1NEET 2021
A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly :
Not quite — the answer is B.
Speed just before impact v = √(2gh) = √200 ≈ 14.1 m/s; it rebounds to the same height, so it leaves with 14.1 m/s upward. Impulse = change in momentum = 2mv = 2 × 0.15 × 14.1 ≈ 4.2 kg m/s. 2.1 kg m/s ignores the reversal of direction.
Q2NEET 2020 · Q160
Two bodies of mass 4 kg and 6 kg are tied to the ends of a massless string. The string passes over a pulley which is frictionless (see figure). The acceleration of the system in terms of acceleration due to gravity (g) is :
Not quite — the answer is C.
For two masses hanging over a frictionless pulley, a = (m₂ − m₁)g/(m₁ + m₂) = (6 − 4)g/10 = g/5. Using only the heavier mass in the denominator, or forgetting the difference of weights, gives the other values.
Q3NEET 2019 · Q2
A mass m is attached to a thin wire and whirled in a vertical circle. The wire is most likely to break when:
Not quite — the answer is C.
At the lowest point the tension must supply the centripetal force and also balance the full weight, T = mg + mv²/r, and the speed is also maximum there, so tension is greatest. At the highest point gravity helps provide the centripetal force, so tension is least.
Q4NEET 2019 · Q30
When an object is shot from the bottom of a long smooth inclined plane kept at an angle 60° with the horizontal, it can travel a distance x₁ along the plane. But when the inclination is decreased to 30° and the same object is shot with the same velocity, it can travel x₂ distance. Then x₁ : x₂ will be:
Not quite — the answer is C.
On a smooth incline the deceleration is g sinθ, so x = u²/(2g sinθ). x₁/x₂ = sin30°/sin60° = (1/2)/(√3/2) = 1 : √3. A steeper incline decelerates the object more, so x₁ must be the smaller distance.
Q5NEET 2019 · Q35
A block of mass 10 kg is in contact against the inner wall of a hollow cylindrical drum of radius 1 m. The coefficient of friction between the block and the inner wall of the cylinder is 0.1. The minimum angular velocity needed for the cylinder to keep the block stationary when the cylinder is vertical and rotating about its axis, will be: (g = 10 m/s²)
Not quite — the answer is C.
The normal force supplies the centripetal force, N = mω²r, and friction must balance weight: μmω²r ≥ mg. ω² = g/(μr) = 10/(0.1 × 1) = 100, so ω = 10 rad/s. √10 rad/s forgets to divide by μ.
Q6NEET 2018 · Q33
Which one of the following statements is incorrect?
Not quite — the answer is D.
The coefficient of friction is a ratio of two forces (friction/normal reaction), so it is dimensionless; the statement that it has dimensions of length is wrong. The other statements are standard properties of friction.
Q7NEET 2017 (Gujarati paper) · Q140
A body initially at rest, breaks up into two pieces of masses 2M and 3M respectively, together having a total kinetic energy E. The piece of mass 2M, after breaking up, has a kinetic energy:
Not quite — the answer is C.
Momentum conservation gives equal and opposite momenta p, and KE = p²/2m, so kinetic energy is inversely proportional to mass. KE(2M) : KE(3M) = 3 : 2, so the 2M piece gets 3E/5. 2E/5 is what you get by wrongly sharing the energy in proportion to mass.
Q8NEET 2017 (Gujarati paper) · Q152
A cyclist on a level road takes a sharp circular turn of radius 3 m (g = 10 m s⁻²). If the coefficient of static friction between the cycle tyres and the road is 0.2, at which of the following speeds will the cyclist not skid while taking the turn?
Not quite — the answer is A.
Friction must supply the centripetal force, so v_max = √(μgr) = √(0.2 × 10 × 3) = √6 ≈ 2.45 m s⁻¹ ≈ 8.8 km h⁻¹. Only 7.2 km h⁻¹ (2 m s⁻¹) is below this limit; 9 km h⁻¹ (2.5 m s⁻¹) is just above it, so the cyclist would skid.
Q9NEET 2016 Phase 1 · Q45
A car is negotiating a curved road of radius R. The road is banked at an angle θ. The coefficient of friction between the tyres of the car and the road is μₛ. The maximum safe velocity on this road is:
Not quite — the answer is B.
With friction acting down the incline at the maximum speed, resolving forces gives v²/(Rg) = (sinθ + μₛcosθ)/(cosθ − μₛsinθ) = (μₛ + tanθ)/(1 − μₛtanθ). So v_max = √(gR(μₛ + tanθ)/(1 − μₛtanθ)); only this form has units of velocity.
Q10AIPMT 2015
A block A of mass m₁ rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m₂ is suspended. The coefficient of kinetic friction between the block and the table is μₖ. When the block A is sliding on the table, the tension in the string is:
Not quite — the answer is D.
a = (m₂ − μₖm₁)g/(m₁ + m₂). For block A: T − μₖm₁g = m₁a, so T = m₁[(m₂ − μₖm₁)g/(m₁ + m₂) + μₖg] = m₁m₂(1 + μₖ)g/(m₁ + m₂). Friction increases the tension compared with the frictionless value.
Q11AIPMT 2014
A system consists of three masses m₁, m₂ and m₃ connected by a string passing over a pulley P. The mass m₁ hangs freely and m₂ and m₃ are on a rough horizontal table (the coefficient of friction = μ). The pulley is frictionless and of negligible mass. The downward acceleration of mass m₁ is: (Assume m₁ = m₂ = m₃ = m)
Not quite — the answer is B.
The driving force is m₁g = mg and friction on the two table masses is 2μmg, while all three masses (3m) accelerate together. a = (mg − 2μmg)/(3m) = g(1 − 2μ)/3.
Q12AIPMT 2014
A balloon with mass 'm' is descending down with an acceleration 'a' (where a < g). How much mass should be removed from it so that it starts moving up with an acceleration 'a'?
Not quite — the answer is D.
Descending: mg − U = ma, so the upthrust U = m(g − a). Rising with mass m′: U − m′g = m′a, so m′ = m(g − a)/(g + a). Mass removed = m − m′ = 2ma/(g + a).
Q13NEET 2013
Three blocks with masses m, 2m and 3m are connected by strings (m at the top, 2m below it and 3m at the bottom). After an upward force F is applied on block m, the masses move upward at constant speed v. What is the net force on the block of mass 2m? (g is the acceleration due to gravity)
Not quite — the answer is A.
All three blocks move at constant velocity, so their acceleration is zero and by Newton's second law the net force on each block, including 2m, is zero. 2mg is only the weight of that block, which is balanced by the string tensions.
Q14NEET 2013
The upper half of an inclined plane of inclination θ is perfectly smooth while the lower half is rough. A block starting from rest at the top of the plane will again come to rest at the bottom, if the coefficient of friction between the block and lower half of the plane is given by
Not quite — the answer is C.
Let the plane length be L. Gain in KE from gravity over the full length, mg sin θ·L, must equal the friction work on the rough lower half, μ mg cos θ·(L/2). This gives μ = 2 tan θ. μ = tan θ would only stop the acceleration on the rough part, not bring the block to rest.
Q15NEET 2013
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m s⁻¹ and the second part of mass 2 kg moves with 8 m s⁻¹ speed. If the third part flies off with 4 m s⁻¹ speed, then its mass is
Not quite — the answer is B.
Initial momentum is zero. The first two momenta are 1 × 12 = 12 and 2 × 8 = 16 kg m s⁻¹ at right angles, giving a resultant √(12² + 16²) = 20 kg m s⁻¹. The third part must carry 20 kg m s⁻¹ in the opposite direction, so m = 20/4 = 5 kg.
Q16AIPMT 2012 Mains
A car of mass m is moving on a level circular track of radius R. If μs represents the static friction between the road and tyres of the car, the maximum speed of the car in circular motion is given by
Not quite — the answer is D.
Static friction supplies the centripetal force: mv²/R ≤ μs·mg, so vmax = √(μs Rg). The mass cancels, so expressions containing m cannot be right.
Q17AIPMT 2012 Prelims
A car of mass 1000 kg negotiates a banked curve of radius 90 m on a frictionless road. If the banking angle is 45°, the speed of the car is
Not quite — the answer is C.
On a frictionless banked road, tan θ = v²/(rg). With θ = 45°, v = √(rg) = √(90 × 10) = 30 m s⁻¹ (g = 10 m s⁻²). The mass of the car does not enter the result.
Q18AIPMT 2011 Mains
A conveyor belt is moving at a constant speed of 2 m/s. A box is gently dropped on it. The coefficient of friction between them is μ = 0.5. The distance that the box will move relative to belt before coming to rest on it, taking g = 10 m s⁻², is
Not quite — the answer is B.
Relative to the belt the box starts at 2 m/s and friction gives it a relative deceleration μg = 5 m/s². Relative distance = u²/2a = 4/10 = 0.4 m. The answer zero ignores that the box slips until it reaches the belt speed.
Q19AIPMT 2011 Prelims
A person of mass 60 kg is inside a lift of mass 940 kg and presses the button on the control panel. The lift starts moving upwards with an acceleration 1.0 m/s². If g = 10 m s⁻², the tension in the supporting cable is
Not quite — the answer is D.
The cable accelerates the lift and the person together (total mass 1000 kg) upward at 1 m/s², so T − (M + m)g = (M + m)a, giving T = 1000 × (10 + 1) = 11000 N. Using only the lift's mass or ignoring the acceleration gives the smaller values.
Q20AIPMT 2011 Prelims
A body of mass M hits normally a rigid wall with velocity V and bounces back with the same velocity. The impulse experienced by the body is
Not quite — the answer is D.
Impulse equals change in momentum: taking the initial direction as positive, Δp = (−MV) − (MV) = −2MV, so its magnitude is 2MV. Zero would only be the change in speed, not in momentum.
Q21AIPMT 2010 Prelims · Q9
A gramophone record is revolving with an angular velocity ω. A coin is placed at a distance r from the centre of the record. The static coefficient of friction is μ. The coin will revolve with the record if
Not quite — the answer is D.
Static friction supplies the centripetal force: mrω² ≤ μmg, so the coin stays on if r ≤ μg/ω². The reversed inequality r ≥ μg/ω² describes when the coin slips off.
Q22AIPMT 2009 Prelims · Q7
A body, under the action of a force F⃗ = 6î − 8ĵ + 10k̂, acquires an acceleration of 1 m/s². The mass of this body must be:
Not quite — the answer is C.
|F| = √(6² + 8² + 10²) = √200 = 10√2 N. With a = 1 m/s², m = F/a = 10√2 kg. Simply adding the components or taking 2√10 is a mistake in the magnitude.
Q23AIPMT 2009 Prelims · Q14
The mass of a lift is 2000 kg. When the tension in the supporting cable is 28000 N, then its acceleration is:
Not quite — the answer is A.
T − mg = ma gives 28000 − 2000 × 10 = 2000a, so a = 4 m s⁻² upwards. The tension exceeds the weight, so the acceleration must be upward; 14 m s⁻² ignores gravity.
Q24AIPMT 2009 Prelims · Q39
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 m s⁻¹ and 2 kg second part moving with a velocity of 8 m s⁻¹. If the third part flies off with a velocity of 4 m s⁻¹, its mass would be:
Not quite — the answer is D.
Momenta of the two pieces are 1 × 12 = 12 and 2 × 8 = 16 kg m s⁻¹ at right angles, with resultant √(12² + 16²) = 20 kg m s⁻¹. The third piece must carry an equal and opposite 20 kg m s⁻¹, so m = 20/4 = 5 kg.
Q25AIPMT 2008 Prelims · Q6
A particle of mass m is projected with velocity v making an angle of 45° with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be
Not quite — the answer is D.
The horizontal momentum mv cos 45° is unchanged, while the vertical component mv sin 45° reverses. The change is 2mv sin 45° = √2 mv. The answer is not zero, because momentum is a vector and its direction has changed even though the speed is the same.
Q26AIPMT 2008 Prelims · Q7
Sand is being dropped on a conveyor belt at the rate of M kg/s. The force necessary to keep the belt moving with a constant velocity of v m/s will be
Not quite — the answer is B.
Each second a mass M of sand must be brought from zero to speed v, so F = v(dm/dt) = Mv newton. The force is not zero: even at constant belt speed, the momentum of the system keeps increasing as sand is added.
Q27AIPMT 2008 Prelims · Q10
A shell of mass 200 g is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is
Not quite — the answer is B.
Momentum conservation: 0.2v = 4V, so the gun recoils with V = v/20. Energy: (1/2) × 0.2 × v² + (1/2) × 4 × (v/20)² = 0.1v² + 0.005v² = 0.105v² = 1050, so v² = 10⁴ and v = 100 m s⁻¹. Ignoring the gun's recoil energy would give about 102 m s⁻¹, so the recoil share must be included.
Q28AIPMT 2007 Prelims · Q17
A block B is pushed momentarily along a horizontal surface with an initial velocity V. If μ is the coefficient of sliding friction between B and the surface, block B will come to rest after a time
Not quite — the answer is D.
Friction gives a deceleration μg, so the block stops when V − μgt = 0, i.e. t = V/(μg). V/g ignores the friction coefficient.
Q29AIPMT 2006 Prelims · Q29
A 0.5 kg ball moving with a speed of 12 m/s strikes a hard wall at an angle of 30° with the wall. It is reflected with the same speed and at the same angle. If the ball is in contact with the wall for 0.25 s, the average force acting on the wall is :
Not quite — the answer is B.
The ball meets the wall at 30° to the wall, so its velocity component normal to the wall is 12 sin 30° = 6 m/s; this component reverses while the parallel component is unchanged. Δp = 2 × 0.5 × 6 = 6 kg m/s and F = 6/0.25 = 24 N. Using cos 30° (treating the angle as measured from the normal) gives about 41.6 N, which is not offered.
Q30AIPMT 2005 Prelims · Q4
A bomb of mass 30 kg at rest explodes into two pieces of masses 18 kg and 12 kg. The velocity of 18 kg mass is 6 m s⁻¹. The kinetic energy of the other mass is :
Not quite — the answer is B.
Momentum conservation from rest: 18 × 6 = 12 × v, so v = 9 m/s. KE = ½ × 12 × 81 = 486 J. 243 J is half the correct value.
How often Laws of Motion came up
Questions from this chapter in each paper on this site.
- NEET 2021: 1
- NEET 2020: 1
- NEET 2019: 3
- NEET 2018: 1
- NEET 2017 (Gujarati paper): 2
- NEET 2016 Phase 1: 1
- AIPMT 2015: 1
- AIPMT 2014: 2
- NEET 2013: 3
- AIPMT 2012 Mains: 1
- AIPMT 2012 Prelims: 1
- AIPMT 2011 Mains: 1
- AIPMT 2011 Prelims: 2
- AIPMT 2010 Prelims: 1
- AIPMT 2009 Prelims: 3
- AIPMT 2008 Prelims: 3
- AIPMT 2007 Prelims: 1
- AIPMT 2006 Prelims: 1
- AIPMT 2005 Prelims: 1
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