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Centre of Mass, Momentum & Collisions — JEE Main Physics MCQs with Solutions
Free JEE Main Physics Centre of Mass, Momentum & Collisions MCQs with step-by-step solutions covering COM of Discrete Mass Systems, COM of Continuous Bodies, COM of Cavity & Truncated Bodies, Motion of COM & Reference Frames, Conservation of Linear Momentum, Impulse & Impulse-Momentum Theorem. 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 — COM of Discrete Mass Systems · hard · numerical
Masses 1 kg, 2 kg, 3 kg are at (0,0), (2,0) and (0,2) m. The coordinates of the COM are:
A. (1, 1) m
B. (2/3, 2/3) m
C. (1, 2/3) m
D. (2/3, 1) m ✓ Correct
Solution: x_cm = (0 + 4 + 0)/6 = 2/3; y_cm = (0 + 0 + 6)/6 = 1.
Q22 — COM of Discrete Mass Systems · hard · numerical
From a system of masses m at each corner of a square of side a, the mass at one corner is doubled (2m). The COM shifts from the centre by:
A. a/4
B. a/5
C. a√2/8
D. a√2/10 towards the doubled corner ✓ Correct
Solution: Shift = (extra mass)·(distance of that corner from centre)/(total) = m·(a/√2)/(5m) = a/(5√2) = a√2/10.
Q23 — COM of Continuous Bodies · hard · numerical
A uniform solid CONE of height 12 cm stands on its base. Its centre of mass is above the base at:
A. 6 cm
B. 8 cm
C. 4 cm
D. 3 cm ✓ Correct
Solution: COM of a solid cone is h/4 above the base = 3 cm (or 3h/4 below the apex).
Q24 — COM of Continuous Bodies · hard · numerical
A rod of length L has density λ = λ₀(1 + x/L). Its COM from x = 0 is at:
A. 4L/9
B. L/2
C. 2L/3
D. 5L/9 ✓ Correct
Solution: M = λ₀(L + L/2) = 3λ₀L/2; ∫xλdx = λ₀(L²/2 + L²/3) = 5λ₀L²/6. x_cm = (5L²/6)/(3L/2) = 5L/9.
Q25 — COM of Cavity & Truncated Bodies · hard · numerical
From a uniform square plate of side 2a, one quadrant (an a×a square) is removed. The COM of the remaining plate shifts from the centre by a distance of:
A. a/6
B. a/3
C. a√2/3
D. a√2/6 ✓ Correct
Solution: The removed quarter (mass M/4) had its centre at (a/2, a/2) — a distance a√2/2 from the plate centre. Shift = (M/4)(a√2/2)/(3M/4) = a√2/6, directed away from the removed corner.
Q26 — COM of Cavity & Truncated Bodies · hard · numerical
From a solid sphere of radius R, a spherical cavity of radius R/2 is scooped out with its centre at R/2 from the sphere centre. The COM of the remainder is displaced from the centre by:
A. R/14 ✓ Correct
B. R/7
C. R/8
D. R/16
Solution: m_cavity = M/8 at d = R/2: shift = (M/8)(R/2)/(7M/8) = R/14, opposite the cavity.
Q27 — Motion of COM & Reference Frames · hard · numerical
A man of mass 60 kg walks 3 m (relative to the boat) towards the shore end of a 120 kg boat on frictionless water. The boat moves back by:
A. 1.5 m
B. 1 m ✓ Correct
C. 2 m
D. 0.75 m
Solution: COM stays fixed: the man moves (3 − x) over the ground while the boat moves x back ⇒ 60(3 − x) = 120x ⇒ x = 1 m. Trap: the man's 3 m is relative to the BOAT, not the ground.
Q28 — Motion of COM & Reference Frames · hard · numerical
A projectile at the top of its path (speed 10 m/s, horizontal) explodes into two equal halves; one comes momentarily to rest. The other half's speed just after is:
A. 20 m/s horizontally ✓ Correct
B. 10 m/s
C. 14.1 m/s
D. 5 m/s
Solution: p conserved at the instant: m×10 = (m/2)×0 + (m/2)v ⇒ v = 20 m/s, still horizontal.
Q29 — Conservation of Linear Momentum · hard · numerical
A shell at rest explodes into two pieces with mass ratio 1 : 3. The ratio of their kinetic energies (light : heavy) is:
A. 1 : 1
B. 1 : 3
C. 9 : 1
D. 3 : 1 ✓ Correct
Solution: Equal/opposite momenta ⇒ KE = p²/2m ∝ 1/m ⇒ 3:1 — the lighter piece takes the lion's share of energy.
Q30 — Conservation of Linear Momentum · hard · numerical
A 1 kg ball moving at 12 m/s splits into two 0.5 kg halves; one moves off at 16 m/s at 90° to the original line. The speed of the other half is:
A. 20 m/s
B. 24 m/s
C. ≈ 28.8 m/s ✓ Correct
D. 16 m/s
Solution: Conserve both components. Perpendicular: 0.5×16 = 8 must be cancelled ⇒ other half v_y = −16 m/s. Along the line: 1×12 = 0.5v_x ⇒ v_x = 24 m/s. Speed = √(24² + 16²) = √832 ≈ 28.8 m/s.