A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictionless air track. Determine the speed of the cart and clay after the collision.
A)zero m/s
B)3 m/s
C)6 m/s
D)9 m/s
E)12 m/s

Question

Quizplus · Accepted Answer

The momentum before the collision is given by:
p_before = m_clay * v_clay = 0.050 kg * 12 m/s = 0.6 kg*m/s
Since the cart is stationary, its momentum is zero: 
p_before = m_cart * v_cart = 0 kg*m/s
After the collision, the clay and cart move together as a system with a common velocity v_final. Therefore, the momentum of the system is: 
p_after = (m_cart + m_clay) * v_final
Applying the law of conservation of momentum, we have: 
p_before = p_after
0.6 kg*m/s = (0.15 kg + 0.050 kg) * v_final
v_final = 3 m/s
Therefore, the speed of the cart and clay after the collision is 3 m/s, which is choice B.

A 0050-Kg Lump of Clay Moving Horizontally at 12 M/s Strikes