Inelastic Collision of a Bullet into a Block

[Zaphia]

Homework Statement

A rifle bullet with mass 8.00g strikes and embeds itself in a block with mass 0.992 kg that rests on a frictionless, horizontal surface and is attached to a coil spring. The impact compresses the spring 15.0 cm. Calibration of the spring shows that a force of 0.750 N is required to compress the spring 0.250 cm. (a) Find the velocity of the block right after impact (b) what was the initial speed of the bullet?

The book lists the answers to be (a) 2.60 m/s and (b) 325 m/s.

Homework Equations

Kenetic Energy = .5mv^2
Hooke's Law: F=kx
m1v1=m2v2

The Attempt at a Solution

Alright, I have tried it 2 different ways (and gotten 2 sets of completely different answers) but I think I'm missing something.

First train of thought:
1. Find the spring constant, k, using the calibration information, gives me a k=300
2. Find the impact of compression by solving F=kx with my compression distance (.15 meters) and 'k' value (F=45Newtons)
3. Find velocity for (a) using KE=(1/2)mv^2 which gives me the velocity=9.49 m/s.
4. Initial speed of bullet for (b) via m1v1=m2v2, Plug values in, giving me a v1=1186.25 m/s

Second train of thought:
1. Momentum of block with bullet=momentum of bullet
so .5V^2=.5(.008)v^2
2. Force constant of spring 'k' = (Force/Compression) = .75/.0025 = 300 Newton/meter (same as previous method)
3. Potential Energy in Spring: .5kx^2=3.375 Joules
4. Kinetic Energy of Block = Potential Energy of Spring, thus (from step 1), .004v^2=3.375 Joules ---> v=29.05 m/s (for part (b))
5. Velocity of Block w/ Bullet V=.008v = .8216 m/s (for part (a)).

Any suggestions would be most useful, thank you :)

 

[andrevdh]

The restoring force of the spring does work on the bullet to stop it, that is the spring's restoring force supplies the stopping force for the bullet. This force is transferred to the bullet via friction (and it is also the same force that the spring is experiencing from the bullet via Newton's third law - in the opposite direction though). The work done by this force is used to change the kinetic energy of the bullet, that is try using the work - kinetic energy theorem. The question arises how one should incorporate the change in kinetic energy of block during the process since it also experiences a change in motion.

 

[ogb p]

Based on your attempts, it seems like you are on the right track. However, there are a few things to consider in order to arrive at the correct answers.

Firstly, in order to find the velocity of the block after impact, you need to take into account the conservation of momentum. This means that the momentum of the bullet before impact is equal to the momentum of the block with the bullet embedded in it after impact. So, using the equation m1v1 = m2v2, you can solve for the velocity of the block after impact (a).

Secondly, when calculating the velocity of the bullet before impact (b), you need to take into account the fact that the bullet is embedded in the block and therefore the mass of the bullet is now part of the block's mass. So, using the same equation as before, m1v1 = m2v2, you can solve for the initial velocity of the bullet.

Lastly, in your second train of thought, you have correctly used the conservation of energy to find the velocity of the block with the bullet embedded in it. However, the velocity you have calculated is the velocity of the block with the bullet, not just the block itself. In order to find the velocity of the block alone, you need to subtract the velocity of the bullet from the total velocity.

I hope this helps clarify the process for solving this problem. Remember to always consider the conservation of momentum and energy in collisions, and to take into account any changes in mass or velocity due to the collision. Keep up the good work!