Question

A 12.0g bullet is fired horizontally into a 650g block that is initially at rest on a frictionless horizontal surface. The initial velocity of the bullet is 450m/s. After the bullet is embedded into the block, the bullet-block system slides along the frictionless surface into a spring having spring constant k=470N/m

a. What is the speed of the block after the bullet once it’s stuck in the block

b. What was the work done on the bullet during the collision

c. Why is the magnitude of the work done on the bullet not the same as the magnitude of the work done on the block

d. Calculate the distance the spring is compressed when the bullet-block system comes to a stop

Answer 1

(a)

by using Conservation of momentum

0.012 * 450 + 0.650 * 0 = ( 0.012 + 0.650 ) * V

V = 8.157 m/s

(b)

Work done on bullet = Change in kinetic energy of bullet

= 1/2 * 0.012 * ( 8.157^2 - 450^2 )

= -1214.607 J

(c)

They are not same as the change in kinetic energies are different.

(d)

from conservation of energy principle,

Kinetic energy of bullet + block ( 1/2 mv^2) = potential energy of spring ( 1/2 kx^2 )

1/2 * ( 0.012 + 0.650 ) * 8.157^2 = 1/2 * 470 * x^2

x = 0.306 m