Conservation of Momentum MCQs

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.