Posted by Jason on .
A(n) 11 g bullet is ﬁred into a(n) 111 g block
of wood at rest on a horizontal surface and
stays inside. After impact, the block slides
14 m before coming to rest.
The acceleration of gravity is 9.8 m/s
If the coeﬃcient of friction between the
surface and the block is 0.7, ﬁnd the speed of
the bullet before impact.
Answer in units of m/s
The kinetic energy of the bullet and block after impact equals the work done against friction.
(1/2)(M+m)V2^2 = (M+m)*g*(0.7)*14m
V2^2 = 2*g*(0.7)(14) = 192 m^2/s^2
V2 = 13.9 m/s
Now use conservation of momentum to determine the velocity of the bullet, V1
m*V1 = (m+M)*V2
V1 = V2*(121/11) = 11 V2 = 152 m/s