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System of Particles and Rotational Motion — JEE Main Physics MCQs with Solutions
Free JEE Main Physics System of Particles and Rotational Motion MCQs with step-by-step solutions covering COM of Discrete Mass Systems, Elastic Collision, COM of Continuous Bodies, Centre of Mass, Torque and Angular Momentum, Inelastic Collision, COM of Cavity & Truncated Bodies. Practise online on Prepizo — no login needed.
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Sample questions with solutions
Q1 — COM of Discrete Mass Systems · easy · theory
The centre of mass of a two-particle system always lies:
A. Outside the line joining them
B. Exactly midway between them
C. Closer to the lighter one
D. On the line joining the particles, closer to the heavier one ✓ Correct
Solution: x_cm = (m₁x₁ + m₂x₂)/(m₁ + m₂); the mass-weighted average sits nearer the larger mass. Midway only for equal masses.
Q2 — COM of Discrete Mass Systems · easy · numerical
Masses 2 kg and 3 kg are at x = 0 and x = 5 m. The centre of mass is at x =
A. 2 m
B. 2.5 m
C. 3 m ✓ Correct
D. 3.5 m
Solution: x_cm = (2×0 + 3×5)/5 = 3 m — closer to the 3 kg mass.
Q3 — COM of Continuous Bodies · easy · theory
The centre of mass of a uniform semicircular RING of radius R lies on the symmetry axis at a distance from the centre of:
A. 4R/3π
B. 2R/π ✓ Correct
C. R/π
D. R/2
Solution: Standard results: semicircular ring 2R/π; semicircular DISC 4R/3π — don't swap them.
Q4 — COM of Continuous Bodies · easy · numerical
A uniform rod of length 2 m has its centre of mass at a distance from one end of:
A. 0.5 m
B. 2/3 m
C. 1 m ✓ Correct
D. 1.5 m
Solution: A uniform rod's COM is at its midpoint: L/2 = 1 m.
Q5 — COM of Cavity & Truncated Bodies · easy · theory
To find the COM of a body with a cavity, the standard method is to treat the cavity as:
A. A positive extra mass
B. A point mass at the edge
C. A superposed NEGATIVE mass of the same shape ✓ Correct
D. Zero mass at the centre
Solution: Full body (positive) + cavity region (negative mass) reproduces the actual object; apply the usual COM formula with the negative term.
Q6 — COM of Cavity & Truncated Bodies · easy · numerical
From a uniform disc of radius R a concentric hole of radius R/2 is cut. The COM of the ring-like remainder is:
A. At R/4 from the centre
B. At R/2
C. Undefined
D. At the original centre ✓ Correct
Solution: Concentric removal keeps full symmetry — COM stays at the centre.
Q7 — Motion of COM & Reference Frames · easy · theory
The centre of mass of a system accelerates only if:
A. A net EXTERNAL force acts on the system ✓ Correct
B. The system rotates
C. Any internal forces act
D. The particles collide
Solution: M a⃗_cm = F⃗_ext. Internal forces cancel in action–reaction pairs and can never move the COM.
Q8 — Motion of COM & Reference Frames · easy · numerical
Two blocks, 2 kg at 6 m/s and 4 kg at 3 m/s, move in the same direction. The velocity of their centre of mass is:
A. 6 m/s
B. 3 m/s
C. 4 m/s ✓ Correct
D. 4.5 m/s
Solution: v_cm = (2×6 + 4×3)/6 = 24/6 = 4 m/s.
Q9 — Conservation of Linear Momentum · easy · theory
The total linear momentum of a system is conserved when:
A. The bodies are rigid
B. Gravity is absent
C. Kinetic energy is conserved
D. The net external force on the system is zero ✓ Correct
Solution: dP⃗/dt = F⃗_ext; zero net external force ⇒ P⃗ constant, whatever the internal forces (collisions, explosions).
Q10 — Conservation of Linear Momentum · easy · numerical
A 40 kg boy on frictionless ice throws a 2 kg ball at 10 m/s. His recoil speed is:
A. 0.05 m/s
B. 0.5 m/s ✓ Correct
C. 2 m/s
D. 5 m/s
Solution: 40v = 2×10 ⇒ v = 0.5 m/s backwards.
Q11 — Impulse & Impulse-Momentum Theorem · easy · theory
Impulse of a force equals:
A. The change in kinetic energy
B. The change in momentum it produces (∫F dt = Δp) ✓ Correct
C. Force × displacement
D. Power × time
Solution: J⃗ = ∫F⃗dt = Δp⃗; its unit N·s ≡ kg·m/s. Graphically, the area under the F–t curve.
Q12 — Impulse & Impulse-Momentum Theorem · easy · numerical
A 0.15 kg cricket ball arrives at 20 m/s and is caught and stopped in 0.1 s. The average force on the hands is:
A. 30 N ✓ Correct
B. 300 N
C. 15 N
D. 3 N
Solution: F = Δp/Δt = 0.15×20/0.1 = 30 N.
Q13 — Elastic Collisions (1D & 2D) · easy · theory
In a perfectly elastic collision, the quantities conserved are:
A. Both linear momentum and kinetic energy ✓ Correct
B. Neither
C. Only momentum
D. Only kinetic energy
Solution: Elastic: p⃗ and KE both conserved. (Momentum is conserved in ALL collisions; KE only in elastic ones.)
Q14 — Elastic Collisions (1D & 2D) · easy · numerical
A 2 kg ball at 6 m/s hits an identical stationary ball head-on, perfectly elastically. The speeds after are:
A. 3 and 3 m/s
B. 0 and 6 m/s ✓ Correct
C. 2 and 4 m/s
D. 6 and 6 m/s
Solution: Equal masses exchange velocities: the incoming ball stops; the struck one leaves at 6 m/s.
Q15 — Inelastic & Perfectly Inelastic Collisions · easy · theory
In a perfectly INELASTIC collision, the two bodies:
A. Stick together and move with a common velocity; KE is lost but momentum is conserved ✓ Correct
B. Exchange velocities
C. Conserve kinetic energy
D. Both stop always
Solution: Sticking = maximum possible KE loss consistent with momentum conservation. Momentum is still exactly conserved.
Q16 — Inelastic & Perfectly Inelastic Collisions · easy · numerical
A 2 kg body at 6 m/s hits a 4 kg body at rest and sticks to it. Their common velocity is:
A. 3 m/s
B. 6 m/s
C. 1.5 m/s
D. 2 m/s ✓ Correct
Solution: v = 2×6/6 = 2 m/s.
Q17 — Coefficient of Restitution & Rebound · easy · theory
The coefficient of restitution e is defined as:
A. (Relative speed of separation)/(relative speed of approach) ✓ Correct
B. The impulse ratio
C. The ratio of final to initial kinetic energy
D. The ratio of the masses
Solution: e = v_sep/v_app along the line of impact; e = 1 elastic, e = 0 perfectly inelastic, 0 < e < 1 real collisions.
Q18 — Coefficient of Restitution & Rebound · easy · numerical
A ball dropped from 5 m rebounds to 3.2 m. The coefficient of restitution is:
A. 0.8 ✓ Correct
B. 0.64
C. 0.36
D. 0.5
Solution: e = √(3.2/5) = √0.64 = 0.8.
Q19 — Variable Mass Systems · easy · theory
The thrust on a rocket is given by:
A. F = v²(dm/dt)
B. F = v_rel(dm/dt), from the momentum of the ejected exhaust ✓ Correct
C. F = mg
D. F = ma always
Solution: Expelling mass at relative speed v_rel carries momentum away at rate v_rel·dm/dt — the reaction is the thrust.
Q20 — Variable Mass Systems · easy · numerical
A rocket ejects gas at 2 kg/s with relative speed 500 m/s. The thrust is:
A. 250 N
B. 500 N
C. 2000 N
D. 1000 N ✓ Correct
Solution: F = v_rel(dm/dt) = 500×2 = 1000 N.
Q21 — Elastic Collision · easy · numerical
A 6 kg sphere moving at 9 m/s makes a head-on elastic collision with a stationary 3 kg sphere. Find the speed of the 3 kg sphere just after the collision.
A. 12 m/s ✓ Correct
B. 9 m/s
C. 6 m/s
D. 3 m/s
Solution: For an elastic collision on a stationary target, v₂ = 2m₁u/(m₁ + m₂) = 2 × 6 × 9 / 9 = 12 m/s. Momentum check: 6 × 9 = 54 = 6 × 3 + 3 × 12.
Q22 — Elastic Collision · easy · numerical
An 8 kg block moving at 5 m/s collides head-on elastically with a stationary 2 kg block. Find the velocity of the 8 kg block after the collision.
A. 5 m/s
B. 8 m/s
C. 2 m/s
D. 3 m/s ✓ Correct
Solution: v₁ = (m₁ - m₂)u/(m₁ + m₂) = (8 - 2) × 5 / 10 = 3 m/s in the original direction; the 2 kg block moves off at 8 m/s.
Q23 — Elastic Collision · easy · numerical
A 5 kg ball moving at 6 m/s strikes a stationary 10 kg block head-on in a perfectly elastic collision. Find the speed of the 10 kg block after the collision.
A. 3 m/s
B. 4 m/s ✓ Correct
C. 2 m/s
D. 6 m/s
Solution: v₂ = 2m₁u/(m₁ + m₂) = 2 × 5 × 6 / 15 = 4 m/s, while the 5 kg ball rebounds at 2 m/s. Momentum: 5 × 6 = 30 = 5 × (-2) + 10 × 4.
Q24 — Elastic Collision · easy · numerical
A ball moves at 12 m/s towards a very massive wall that is itself moving towards the ball at 3 m/s. If the collision is perfectly elastic, find the speed of the ball after it bounces off.
A. 12 m/s
B. 15 m/s
C. 18 m/s ✓ Correct
D. 21 m/s
Solution: Relative to the wall the ball approaches at 12 + 3 = 15 m/s and leaves at 15 m/s, so in the ground frame its speed is 15 + 3 = 18 m/s (equivalently u + 2V = 12 + 2 × 3).
Q25 — Elastic Collision · easy · numerical
Ball A moving at 10 m/s strikes an identical stationary ball B head-on elastically; B then strikes an identical stationary ball C head-on elastically. Find the final speed of ball C.
A. 5 m/s
B. 0 m/s
C. 20 m/s
D. 10 m/s ✓ Correct
Solution: In an elastic collision between equal masses the velocities are exchanged: A stops and B moves at 10 m/s, then B stops and C moves at 10 m/s.
Q26 — Elastic Collision · easy · numerical
Two identical 3 kg balls move towards each other with speeds 10 m/s and 4 m/s and collide head-on elastically. Find the speed, after the collision, of the ball that was moving at 4 m/s.
A. 4 m/s
B. 7 m/s
C. 10 m/s ✓ Correct
D. 6 m/s
Solution: Equal masses exchange velocities in a head-on elastic collision, so the 4 m/s ball moves off at 10 m/s (and the other at 4 m/s). Momentum: 3 × 10 - 3 × 4 = 3 × 10 - 3 × 4.
Q27 — Inelastic Collision · easy · numerical
A 4 kg body moving at 15 m/s collides with a stationary 6 kg body and the two stick together. Find their common velocity after the collision.
A. 6 m/s ✓ Correct
B. 9 m/s
C. 4 m/s
D. 15 m/s
Solution: Momentum conservation: 4 × 15 = (4 + 6)v, so v = 60/10 = 6 m/s.
Q28 — Inelastic Collision · easy · numerical
A 6 kg trolley moving at 10 m/s catches up with a 4 kg trolley moving at 5 m/s in the same direction and couples with it. Find the common velocity of the pair.
A. 5 m/s
B. 8 m/s ✓ Correct
C. 7.5 m/s
D. 10 m/s
Solution: v = (6 × 10 + 4 × 5)/(6 + 4) = 80/10 = 8 m/s.
Q29 — Inelastic Collision · easy · numerical
An 8 kg body moving at 6 m/s towards the east collides head-on with a 4 kg body moving at 12 m/s towards the west, and they stick together. Find their velocity after the collision.
A. 2 m/s east
B. 4 m/s west
C. 0 m/s (they come to rest) ✓ Correct
D. 6 m/s east
Solution: Total momentum = 8 × 6 - 4 × 12 = 48 - 48 = 0, so the combined 12 kg mass is at rest after the collision.
Q30 — Inelastic Collision · easy · numerical
A 5 kg body moving at 8 m/s meets a 3 kg body moving at 8 m/s in the opposite direction; they stick together on impact. Find their common velocity.
A. 1 m/s in the direction of the 3 kg body
B. 8 m/s in the direction of the 5 kg body
C. 4 m/s in the direction of the 5 kg body
D. 2 m/s in the direction of the 5 kg body ✓ Correct
Solution: v = (5 × 8 - 3 × 8)/(5 + 3) = 16/8 = 2 m/s, along the initial motion of the heavier (5 kg) body.