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 23.9 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

Answer 1

Answer: 0.43 m

Step-by-step explanation:

Given

mass of bullet
`m=10\ gm\approx 0.01\ kg`

mass of block
`M=1.95\ kg`

The Force constant of spring is
`k=23.9\ N/m`

Speed of bullet is
`u=300\ m/s`

Conserving the energy i.e. kinetic energy of the bullet and box is converted into Elastic potential energy of spring

`\Rightarrow (1)/(2)(M+m)v^2=(1)/(2)kx^2`

Conserving linear momentum

`\Rightarrow mu=(M+m)v\\\\\Rightarrow v=(mu)/(M+m)`

Put the value of
`v` we get

`\Rightarrow x=mu\sqrt{(1)/(k(M+m))}\\\Rightarrow x=0.01* 300\sqrt{(1)/(23.9(0.01+1.95))}\\\Rightarrow x=3\sqrt{(1)/(46.844)}=(3)/(6.844)=0.43\ m`

Thus, spring will be compressed to a distance of
`0.43\ m`