# Homework Help: Determining the speed of the bullet after a collision

1. Oct 16, 2012

### lendav_rott

1. The problem statement, all variables and given/known data
A bullet with a mass of 4g flies horizontally and collides with a ball of 3kg which is hanging down attached to a hawser with the length of 2m. What was the bullet's speed before the collision if the hawser tilts 15 degrees away from its original position

2. Relevant equations

3. The attempt at a solution
Ok I can say the distance the impact forces the ball to move.
d = r * α where r is the length of the hawser and α is the change of angle.

Before I go on, am I doing the degrees-to-radians thing right?

So d = 2m * pi/12 ≈ 0.5236 m

But what comes next? Does it have to do with the bullet's kinetic energy being strong enough to make the ball move 0.5236m? I'm confused :/

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2. Oct 16, 2012

### HallsofIvy

We are not told what happens to the bullet after the collisions so there is enough information here only if we assume this is a "perfectly inelastic" collision. That is the two pieces, the bullet lodges in the ball and the two form an object of mass 3.004 kg (3004 g). The bullet's kinetic energy before the collision is not relevant because energy is not conserved in an inelastic collision. However, momentum is conserved. Assume the bullets speed before the collision is v. The bullets momentum before the collision is .004v and the ball-bullet together after the collision have momentum 4.004u and we those must be equal, 4.004u= .004v, so u= (.004/4.004)v. Knowing that, you can find the kinetic energy of the ball-bullet mass. Now you can use "conservation of energy". That kinetic energy must be equal to the potential energy of the ball-bullet mass at the highest point in the swing: (4.004)gh.

3. Oct 16, 2012

### lendav_rott

Ok so I calculated the change in height of the ball, let it be Δx = 0.06815m

According to conservation of momentum:
m1v1 + m2v2 + (m1+m2)v3 = 0 - these are vectors, how do I put the vector arrow on the speeds?
m1, v1 - mass of the bullet and speed of the bullet
m2, v2 - mass of the ball and speed of the ball
v3 - post-collision speed of the ball-bullet mass

The ball stays still, so its momentum is 0, therefore:

m1v1 = (m1+m2)v3
v3 = m1v1/(m1+m2)

The maximum change in the ball-bullet potential energy at Δx is:

ΔEp = (m1+m2)gΔx

and will have to be equal to the kinetic energy of the ball-bullet mass, which is:

Ek = (m1+m2)v3²/2 = (m1+m2)(m1v1)²/2(m1+m2

So after a bit of simplification here n there I got to:

v1 = (m1+m2)*√2gΔx /m1 ≈ 868 m/s
Is it realistic? It's like 2.5 times faster than the speed of sound.

E: I would really like to know how write the calculus part as a proper calculation , where division is separated by the line etc :/ this is confusing

Last edited: Oct 16, 2012