A4.30 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?

(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.

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

u=13

v=30

s/t=u+1\2at

=13+431/2

=13+265.5

=278.5

probably not if net force is zero