Question

A bullet of mass m is fired into a block of mass M that is at rest. The block, with the bullet embedded, slides distance d across a horizontal surface. The coefficient of kinetic friction is ?_k.

What is the speed of a 9.0 g bullet that, when fired into a 8.0 kg stationary wood block, causes the block to slide 5.6 cm across a wood table? Assume that ?k=0.20.

Answer 1

Given  

mass of bullet m = 9 g

mass of block M = 8 kg

distancemoved (m+M) is s = 5.6 cm

coefficient of kinetic friction is mue_k = 0.20

due to friction the bullet and block system velocity goes on decrease from t=0s

so the work done by the friction force is  

Ff = mue_k *(m+M)g cos 0 = 0.2*(0.009+8)*9.8cos 0 N = 15.69764 N

the work done by Ff is = Ff*s = 15.69764 *0.056 J = 0.87906784 J

this is equal to teh change in kientic energy  

0.5*(m+M)*V^2 = 0.87906784 J

0.5*(0.009+8)*V^2 = 0.87906784

V = 0.468529 m/s

this is final speed  

from conservation of moemntum  

m*u = (m+M)V

u = (m+M)V /m

u = (0.009+8)*0.468529 / (0.009) m/s

u = 416.939 m/s

so the speed of the bullet is u = 413.939 m/s