Question

a bullet of mass 0.005kg moving with a speed of 200m/s enters a heavy wooden block and is stopped after a distance of 50 cm. what is the average force exerted by the block on the bullet?

Answer 1

answer : 200N

explanation : according to Newton's 2nd law, F = ma

so, at first we have to find out retardation of bullet in wooden block.

initial speed of bullet, u = 200m/s

final speed of bullet , v = 0

distance travelled by bullet, s = 50cm =0.5m

use formula, v² = u² + 2as

or, 0 = (200)² + 2a(0.5)

or, a = -40000 m/s² [ here negative sign indicates bullet is retarding in the wooden block. ]

now, force exerted by bullet in the block = ma = 0.005 kg × -40000 m/s²

= -200N

from Newton's 3rd law,

force exerted by block on the bullet = - force exerted by bullet on the block.

= -(-200) N = 200N

Answer 2

Explanation:

Hope it helps and also do not forget to try it yourself before you see the answer.