[figure] Consider two blocks A and B of masses m1 = 10 kg and m2 = 5 kg that are placed on a frictionless table. The block A moves with a constant speed v = 3 m/s towards the block B kept at rest. A spring with spring constant k = 3000 N/m is attached with the block B as shown in the figure. After the collision, suppose that the blocks A and B, along with the spring in constant compression state, move together, then the compression in the spring is, (Neglect the mass of the spring)
- 0.3 m
- 0.1 m
- 0.4 m
- 0.2 m
Answer
(2) 0.1 m
Solution
Momentum is conserved (no external horizontal force):
m1v = (m1 + m2)V
⇒ 10 3 = 15V ⇒ V = 2 m/s
Kinetic energy lost = energy stored in the spring:
12m1v2 - 12(m1 + m2)V2 = 12kx2
⇒ 45 - 30 = 12 3000 x2
⇒ x2 = 151500 = 0.01
∴ x = 0.1 m
Answer: (2)