Conservation of MomentumPractice Questions & MCQs
Practice Conservation of Momentum questions from Force and Laws of Motion (Class 9 Science) - important MCQs with answers and step-by-step solutions. Solve the sample questions below, then practice the full set free on the RankByte app.
Conservation of MomentumQuiz - Solve & Score
Q1. A loaded trolley of mass 3 kg moving at 2 m/s catches up with and locks onto a stationary trolley of mass 6 kg. If momentum is conserved, what is the common velocity with which the coupled trolleys then move together?
- A.2 m/s
- B.1 m/s
- C.0.6667 m/s
- D.0.6 m/s
Answer: C. 0.6667 m/s
Total momentum before = m1*u1 + m2*u2 = 3*2 + 6*0 = 6 kg m/s. After the collision the masses move together with mass (m1+m2) = 9 kg. By conservation v = total momentum / total mass = 6 / 9 = 0.6667 m/s.
Q2. A 4 kg lump of wet clay travelling at 2 m/s strikes and sticks to a 6 kg block resting on a frictionless table. If momentum is conserved, what is the common velocity with which the clay and block move off as one?
- A.0.8 m/s
- B.0.7273 m/s
- C.2 m/s
- D.1.3333 m/s
Answer: A. 0.8 m/s
Total momentum before = m1*u1 + m2*u2 = 4*2 + 6*0 = 8 kg m/s. After the collision the masses move together with mass (m1+m2) = 10 kg. By conservation v = total momentum / total mass = 8 / 10 = 0.8 m/s.
Q3. On smooth ice a 5 kg sledge sliding at 2 m/s runs into a 6 kg sledge that is at rest, and the two hook together. If momentum is conserved, what is the common velocity with which the joined sledges glide on?
- A.0.9091 m/s
- B.2 m/s
- C.1.6667 m/s
- D.0.8333 m/s
Answer: A. 0.9091 m/s
Total momentum before = m1*u1 + m2*u2 = 5*2 + 6*0 = 10 kg m/s. After the collision the masses move together with mass (m1+m2) = 11 kg. By conservation v = total momentum / total mass = 10 / 11 = 0.9091 m/s.
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