Question

A 4.30 g bullet moving at 943 m/s strikes a 730 g wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to 484 m/s. (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

Answer 1

Answer:

(a)2.7 m/s

(b) 5.52 m/s

Explanation:

The total of the system would be conserved as no external force is acting on it.

Initial momentum = final momentum

⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)

⇒ 730 ×v = (4054.9 - 2081.2) =1973.7

⇒v=2.7 m/s

Thus, the resulting speed of the block is 2.7 m/s.

(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

$V_{COM} = \frac{m_b}{m_b+m_{bl}}v_{bi}=\frac{4.30}{4.30+730}\times 943 m/s = 5.52 m/s$

Thus, the speed of the bullet-block center of mass is 5.52 m/s.