Question

A 4.00 g bullet moving at 280 m/s enters and stops in an initially stationary 3.40 kg wooden block on a horizontal frictionless surface.

What's the speed of the bullet/ block combination? 0.33 m/s

What fraction of the bullet's kinetic energy was lost in this perfectly inelastic collision?

How much work was done in stopping the bullet?

If the bullet penetrated 5.00 cm into the wood, what was the average stopping force?

Answer 1

a] By momentum conservation:

Final momentum = initial momentum

(m+M)V = mu where m and M are masses of bullet and Block, u is initial speed of bullet, V is final speed of combination,

(0.004+3.4)V = 0.004*280

V = 0.004*280/ (0.004+3.4)

= 0.329 m/s

b] fraction of KE lost = [initial KE-Final KE]/initial KE = [0.5*mu^2-0.5(M+m)V^2]/0.5*mu^2

= 1 - (M+m)V^2]/mu^2

= 1- (0.004+3.4)*0.329^2/[0.004*280^2]

= 0.9988

c] work done = change in KE = 0.5mu^2 -0.5mV^2 = 0.5*0.004*280^2-0.5*(0.004+3.4)*0.329^2

= 157 J

d] stopping force = Work/d= 157/0.05 = 3140 N