Conservation of Momentum — Homi Bhabha Class 9 Physics MCQs with Solutions
Free Homi Bhabha Class 9 Physics Conservation of Momentum MCQs with step-by-step solutions (10 questions). Part of Laws of Motion & Force. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Conservation of Momentum · hard · numerical
A gun of mass 4 kg fires a bullet of mass 0.02 kg at 100 m/s. The recoil velocity of the gun is:
A. 5 m/s
B. 2 m/s
C. 0.5 m/s ✓ Correct
D. 0.05 m/s
Solution: By conservation of momentum $0 = m_g v_g + m_b v_b \Rightarrow v_g = \dfrac{0.02 \times 100}{4} = 0.5$ m/s (opposite to the bullet).
Q2 — Conservation of Momentum · medium · theory
The recoil of a gun and the forward motion of a rocket are both explained by:
A. conservation of energy
B. the law of friction
C. conservation of momentum ✓ Correct
D. inertia of rest
Solution: In each case, the momentum gained by the ejected mass equals the momentum gained by the system in the opposite direction.
Q3 — Conservation of Momentum · hard · numerical
A 2 kg trolley moving at 3 m/s collides and sticks to a stationary 1 kg trolley. Their common velocity after collision is:
A. 3 m/s
B. 6 m/s
C. 2 m/s ✓ Correct
D. 1.5 m/s
Solution: Momentum conserved: $2 \times 3 = (2+1)v \Rightarrow v = \dfrac{6}{3} = 2$ m/s.
Q4 — Conservation of Momentum · medium · theory
The total momentum of a system is conserved only when:
A. no net external force acts on the system ✓ Correct
B. the bodies have equal mass
C. friction is present
D. the system is at rest
Solution: Momentum is conserved for an isolated system, i.e. when the net external force is zero.
Q5 — Conservation of Momentum · hard · numerical
A stationary bomb of mass 6 kg explodes into two equal pieces. If one piece moves at 4 m/s, the other moves at:
A. 4 m/s in the opposite direction ✓ Correct
B. 8 m/s in the opposite direction
C. 4 m/s in the same direction
D. 2 m/s in the opposite direction
Solution: Initial momentum is zero; with equal masses the two pieces must move with equal speed in opposite directions, so 4 m/s oppositely.
Q6 — Conservation of Momentum · medium · numerical
Two bodies of masses 2 kg and 3 kg move towards each other with momenta of equal magnitude 6 kg·m/s. The total momentum of the system is:
A. 0 kg·m/s ✓ Correct
B. 6 kg·m/s
C. 12 kg·m/s
D. 5 kg·m/s
Solution: Equal and opposite momenta cancel, giving a total system momentum of zero.
Q7 — Conservation of Momentum · easy · theory
When no net external force acts on a system, its total momentum:
A. keeps decreasing steadily
B. remains constant ✓ Correct
C. keeps increasing steadily
D. is always zero
Solution: The law of conservation of momentum states that in the absence of a net external force the total momentum of a system stays constant.
Q8 — Conservation of Momentum · medium · numerical
A gun of mass 4 kg fires a bullet of mass 0.02 kg with a velocity of 200 m/s. The recoil velocity of the gun is:
A. 1 m/s ✓ Correct
B. 0.5 m/s
C. 10 m/s
D. 4 m/s
Solution: By conservation of momentum, 0.02 × 200 = 4 × v, so v = 4/4 = 1 m/s (in the opposite direction to the bullet).
Q9 — Conservation of Momentum · medium · numerical
A 2 kg ball moving at 3 m/s strikes a stationary 1 kg ball and the two stick together and move off together. Their common velocity is:
A. 2 m/s ✓ Correct
B. 3 m/s
C. 1.5 m/s
D. 6 m/s
Solution: Total momentum before = 2 × 3 + 1 × 0 = 6 kg m/s. After sticking, total mass = 3 kg, so common velocity = 6/3 = 2 m/s.
Q10 — Conservation of Momentum · hard · theory
A rocket rises by ejecting hot gases downward at high speed. The forward motion of the rocket is best explained by the principle of:
A. conservation of momentum ✓ Correct
B. inertia of rest
C. conservation of energy
D. the rocket pushing against the surrounding air
Solution: The gases carry momentum downward, so the rocket must gain an equal amount of momentum upward to keep the total constant; this is conservation of momentum, which is why rockets also work in the vacuum of space where there is no air to push against.