Question

A 10.0-g bullet is fired into, and embeds itself in, a 1.95-kg block attached to a spring with a force constant of 16.6 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block and the block slides on a frictionless surface? Note: You must use conservation of momentum in this problem because of the inelastic collision between the bullet and block.

Answer 1

Distance the spring is compressed, x = 0.52 m

Step-by-step explanation:

Given :

Mass of the bullet, m = 10 g = 0.01 kg

Mass of the block, M = 1.95 kg

spring force constant, k = 16.6 N/m

Distance the spring is compressed =x

Speed of the bullet, v = 300 m/s

Speed of the block = V

Therefore we know that according to the law of conservation of momentum,

m.v = ( m+M )V

or
`V = (m* v)/(m+M)`

`V = (0.01* 300)/(0.01+1.95)`

= 1.53 m/s

Now according to the law of conservation of momentum,

`(1)/(2)* M* V^(2) =(1)/(2)* k* x^(2)`

`x = \sqrt{(\left ( M+m \right ).V^(2))/(k)}`

`x = \sqrt{(\left ( M+m \right ))/(k)}* V`

`x = \sqrt{(\left ( 1.95+0.01 \right ))/(16.6)}* 1.53`

`x = 0.52` m

Thus, distance the spring is compressed, x = 0.52 m