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Centre of Mass, Momentum & Collisions — NEET Physics MCQs with Solutions
Free NEET 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. Closer to the lighter one
B. Exactly midway between them
C. Outside the line joining them
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. 3 m ✓ Correct
B. 3.5 m
C. 2 m
D. 2.5 m
Solution: x_cm = (2×0 + 3×5)/5 = 3 m — closer to the 3 kg mass.
Q3 — COM of Discrete Mass Systems · easy · numerical
The separation of H and Cl in an HCl molecule is r; Cl is ~35.5 times heavier than H. The COM lies:
A. Very close to H
B. Outside the molecule
C. At the middle
D. Very close to the Cl atom (≈ r/36.5 from Cl) ✓ Correct
Solution: x from Cl = m_H r/(m_H + m_Cl) = r/36.5 — practically on the chlorine atom.
Q4 — 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. 2R/π ✓ Correct
B. 4R/3π
C. R/2
D. R/π
Solution: Standard results: semicircular ring 2R/π; semicircular DISC 4R/3π — don't swap them.
Q5 — 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. 1 m ✓ Correct
C. 2/3 m
D. 1.5 m
Solution: A uniform rod's COM is at its midpoint: L/2 = 1 m.
Q6 — COM of Continuous Bodies · easy · numerical
The centre of mass of a uniform circular RING lies:
A. At its geometric centre, where there is no mass ✓ Correct
B. On the rim
C. R/2 above the centre
D. At 2R/π from the centre
Solution: By symmetry the COM is the centre — an empty point, showing the COM need not lie within the material.
Q7 — 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.
Q8 — 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. At the original centre ✓ Correct
D. Undefined
Solution: Concentric removal keeps full symmetry — COM stays at the centre.
Q9 — COM of Cavity & Truncated Bodies · easy · numerical
A uniform metre stick is cut at the 40 cm mark and the shorter piece is discarded. The COM of the remaining 60 cm piece (from its own left end at 40 cm) is at the stick mark:
A. 75 cm
B. 60 cm
C. 70 cm ✓ Correct
D. 80 cm
Solution: The 60 cm piece spans 40–100 cm; its midpoint is at 70 cm.
Q10 — Motion of COM & Reference Frames · easy · theory
The centre of mass of a system accelerates only if:
A. The system rotates
B. The particles collide
C. Any internal forces act
D. A net EXTERNAL force acts on the system ✓ Correct
Solution: M a⃗_cm = F⃗_ext. Internal forces cancel in action–reaction pairs and can never move the COM.
Q11 — 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. 4.5 m/s
B. 3 m/s
C. 6 m/s
D. 4 m/s ✓ Correct
Solution: v_cm = (2×6 + 4×3)/6 = 24/6 = 4 m/s.
Q12 — Motion of COM & Reference Frames · easy · numerical
Two skaters, 40 kg and 60 kg, at rest push off each other on smooth ice. Their centre of mass:
A. Remains at rest where it was ✓ Correct
B. Moves towards the lighter skater
C. Oscillates
D. Moves towards the heavier skater
Solution: The pushes are internal; with no net external horizontal force, v_cm stays zero.
Q13 — Conservation of Linear Momentum · easy · theory
The total linear momentum of a system is conserved when:
A. Kinetic energy is conserved
B. The net external force on the system is zero ✓ Correct
C. The bodies are rigid
D. Gravity is absent
Solution: dP⃗/dt = F⃗_ext; zero net external force ⇒ P⃗ constant, whatever the internal forces (collisions, explosions).
Q14 — 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. 0.5 m/s ✓ Correct
B. 0.05 m/s
C. 2 m/s
D. 5 m/s
Solution: 40v = 2×10 ⇒ v = 0.5 m/s backwards.
Q15 — Conservation of Linear Momentum · easy · numerical
A 100 kg gun fires a 1 kg shell at 200 m/s. The gun's recoil speed is:
A. 2 m/s ✓ Correct
B. 20 m/s
C. 0.2 m/s
D. 4 m/s
Solution: v = 1×200/100 = 2 m/s.
Q16 — Impulse & Impulse-Momentum Theorem · easy · theory
Impulse of a force equals:
A. Force × displacement
B. Power × time
C. The change in kinetic energy
D. The change in momentum it produces (∫F dt = Δp) ✓ Correct
Solution: J⃗ = ∫F⃗dt = Δp⃗; its unit N·s ≡ kg·m/s. Graphically, the area under the F–t curve.
Q17 — 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. 15 N
B. 30 N ✓ Correct
C. 3 N
D. 300 N
Solution: F = Δp/Δt = 0.15×20/0.1 = 30 N.
Q18 — Impulse & Impulse-Momentum Theorem · easy · numerical
A 60 kg athlete lands from a jump and stops his 6 m/s fall in 0.3 s by bending his knees. The average force on him is:
A. 360 N
B. 3600 N
C. 1200 N ✓ Correct
D. 120 N
Solution: F = Δp/t = 60×6/0.3 = 1200 N (plus his weight, if asked for the ground reaction).
Q19 — Elastic Collisions (1D & 2D) · easy · theory
In a perfectly elastic collision, the quantities conserved are:
A. Only momentum
B. Neither
C. Only kinetic energy
D. Both linear momentum and kinetic energy ✓ Correct
Solution: Elastic: p⃗ and KE both conserved. (Momentum is conserved in ALL collisions; KE only in elastic ones.)
Q20 — 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. 2 and 4 m/s
B. 0 and 6 m/s ✓ Correct
C. 6 and 6 m/s
D. 3 and 3 m/s
Solution: Equal masses exchange velocities: the incoming ball stops; the struck one leaves at 6 m/s.
Q21 — Elastic Collisions (1D & 2D) · easy · numerical
A moving ball strikes an identical stationary ball elastically but NOT head-on (oblique, smooth). The angle between their final velocities is:
A. 180°
B. 60°
C. 45°
D. 90° ✓ Correct
Solution: For equal masses in an oblique elastic collision the final velocities are perpendicular — a classic result provable from p and KE conservation.
Q22 — Inelastic & Perfectly Inelastic Collisions · easy · theory
In a perfectly INELASTIC collision, the two bodies:
A. Conserve kinetic energy
B. Stick together and move with a common velocity; KE is lost but momentum is conserved ✓ Correct
C. Both stop always
D. Exchange velocities
Solution: Sticking = maximum possible KE loss consistent with momentum conservation. Momentum is still exactly conserved.
Q23 — 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. 2 m/s ✓ Correct
B. 1.5 m/s
C. 3 m/s
D. 6 m/s
Solution: v = 2×6/6 = 2 m/s.
Q24 — Inelastic & Perfectly Inelastic Collisions · easy · numerical
A 3 kg lump of clay at 4 m/s hits a 1 kg block at rest and sticks. The common speed is:
A. 3 m/s ✓ Correct
B. 2 m/s
C. 1 m/s
D. 4 m/s
Solution: v = 12/4 = 3 m/s.
Q25 — 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 ratio of final to initial kinetic energy
C. The impulse ratio
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.
Q26 — Coefficient of Restitution & Rebound · easy · numerical
A ball dropped from 5 m rebounds to 3.2 m. The coefficient of restitution is:
A. 0.36
B. 0.5
C. 0.8 ✓ Correct
D. 0.64
Solution: e = √(3.2/5) = √0.64 = 0.8.
Q27 — Coefficient of Restitution & Rebound · easy · numerical
A ball dropped from 1.8 m rebounds with e = 2/3. The rebound height is:
A. 1.2 m
B. 0.8 m ✓ Correct
C. 0.6 m
D. 0.9 m
Solution: h₁ = e²h = (4/9)×1.8 = 0.8 m.
Q28 — Variable Mass Systems · easy · theory
The thrust on a rocket is given by:
A. F = v_rel(dm/dt), from the momentum of the ejected exhaust ✓ Correct
B. F = ma always
C. F = v²(dm/dt)
D. F = mg
Solution: Expelling mass at relative speed v_rel carries momentum away at rate v_rel·dm/dt — the reaction is the thrust.
Q29 — Variable Mass Systems · easy · numerical
A rocket ejects gas at 2 kg/s with relative speed 500 m/s. The thrust is:
A. 2000 N
B. 500 N
C. 250 N
D. 1000 N ✓ Correct
Solution: F = v_rel(dm/dt) = 500×2 = 1000 N.
Q30 — Variable Mass Systems · easy · numerical
Water leaves a fire hose at 20 kg/s with speed 30 m/s. The reaction force on the hose is:
A. 60 N
B. 600 N ✓ Correct
C. 300 N
D. 150 N
Solution: F = v(dm/dt) = 30×20 = 600 N.