Bullet, Block and Spring

## Question:

A 20 gram bullet is fired horizontally t 1200 m/s into a 300 gram
block which rests on a smooth surface. The back of the block
rests against a spring which is in a relaxed state. The spring
has stiffness k=200N/m. Determine the distance the block moves in
compressing the spring after the bullet hits the block.
## Answer:

Since the collision between the bullet and the block is
inelastic, the sum of kinetic and potential energy is not
conserved. Some of it ends up as heat in the block and the
bullet. Momentum will be conserved though. At the instant the
bullet hits the block it momentum is transfered to the bullet
block combination. This momentum will establish the velocity of
the bullet block combination over that first tiny delta x, before
the spring is compressed significantly.
Initially all the momentum is in the bullet at .02kg times 1200m/s = 24 kgm/s. As soon as the
bullet becomes one with the block, the mass is .320 kg so the
initial block velocity is 24 kgm/s / .320 kg =
75 m/s. The kinetic energy of the block-bullet combination
is 1/2*.32*75^{2} = 900 Joules.
The spring will compress until this kinetic energy is transfered
into potential energy of the spring. The potential energy of a
spring is given by pe =
1/2*k*x^{2} where x is the amount of compression.
So pe=1/2*200*x^{2}=900 Joules
when the block stops compressing the spring. This gives us x^{2}=1800/200 = 9, so x=3m.

This information is brought to you by M. Casco Associates, a
company dedicated to helping humankind reach the stars through
understanding how the universe works. My name is James D. Jones.
If I can be of more help, please let me know.

JDJ