Cart A, with a mass of 0.2 kg, travels on a horizontal air track at 3 m/s and hits cart B, which has a mass of 0.4 kg and is initially at rest. After the collision the center of mass of the two cart system has a speed in m/s of:
A) zero
B) 0.33 m/s
C) 2.3 m/s
D) 2.5
E) 5.0 m/s

Question

Quizplus · Accepted Answer

Before the collision, the momentum of cart A is: p1 = m1*v1 = 0.2 kg * 3 m/s = 0.6 kg*m/s
Cart B is at rest, so its momentum is p2 = m2*0 = 0
The initial total momentum of the system is p_initial = p1 + p2 = 0.6 kg*m/s
After the collision, the carts stick together and move with a common speed v_final. By conservation of momentum, we have:
p_initial = p_final
0.6 kg*m/s = (m1 + m2)*v_final
0.6 kg*m/s = 0.6 kg * v_final
v_final = 1 m/s
Therefore, the center of mass of the two cart system has a speed of 1 m/s after the collision. However, since the initial speed of the center of mass was zero (cart B was at rest), the change in center of mass velocity is:
delta_v_com = v_final - 0 = 1 m/s
Since the mass of the system is m1 + m2 = 0.2 kg + 0.4 kg = 0.6 kg, the velocity of the center of mass changes by delta_v_com / (0.6 kg), which gives us:
delta_v_com / (m1 + m2) = 1 m/s / 0.6 kg = 0.33 m/s
Therefore, the answer is B) 0.33 m/s.

Cart A, with a Mass of 0