# Ballistic pendulum question

1. Dec 7, 2012

### mesa

This is a ballistic pendulum promlem, here is the scenerio that is outlined in my physics book:

We have a wooden block of mass 1.20kg hanging from a massless rope 1.5m in length. A bullet of mass .01kg is fired at the block. The collision is inelastic but thermal losses from impact are unknown. The block travels through to a maximum θ=40o. What is the initial velocity (Vib) of the bullet?

mb=.01kg

Mw=1.20kg

Lrope=1.5m

θ=40

Since theta is a known we can use this to find the energy in the system excluding our unknown thermal energy losses; @ max height we get a PE of:

Ug=(mb+Mw)χgχ(L-Lχcos(θ))

This number can now be used to calculate velocity at the bottom of the swing when all PE has been converted to KE like so:

KE=PE

(1/2)χ(mb+Mw)χv^2=(mb+Mw)χgχ(L-Lχcos(θ))

solve for v:

v=[2(mb+Mw)χgχ(L-Lcos(θ))/((mb+Mw)]^(1/2)

v=[2χgχ(L-Lχcos(θ))]^(1/2)

v=2.62m/s

Now this is where I am having an issue, this is the remainder of the solution shown in the book:

Coservation of linear moentum Pf=Pi

(Mw+mb)χv=Mwχ(Viw)+mbχ(Vib)

Solve for Vib:

Vib=vχ(Mw+mb)/mb

giving our value of Vib=320m/s

Here is where I am having issues; lets change the scenerio slightly, same exact experiment however with two different types of wood. Now let us say the thermal losses in the system of each type of wood and the bullet is slightly different. The first experiment will use pine and lets say it behaves identically to the problem we have 'solved' above. A great deal of the initial kinetic energy of the bullet is lost to thermal energy in our piece of pine resulting in a small theta of 40 degress.

Now lets say the other experiment is identical however the wood is now oak (same mass). If oak/bullet system has a different displacement of thermal energy then more(or less depending on the situation) of the bullets initial KE is put into the system as work resulting in a larger(or smaller, once again depending on thermal losses) theta. Now as theta is increased(or decreased) we know our final velocity will be larger(or smaller), but the book says:

Vib=vχ(Mw+mb)/mb

This means that the Vib(velocity of our bullet) is a function based on the velocity (v) but our v is now a larger(or smaller) value for the oak/bullet system than it is for pine/bullet system so somehow by changing the type of wood (according to the books arguement) results in a change in the velocity (Vib)of an identical bullet from an identical gun.

The only way I see this working is if the thermal losses are identical regardless of the materials that the bullet and block are made of, which would be quite remarkable.

2. Dec 7, 2012

### TSny

Seems remarkable, but your conclusion is true. For a completely inelastic collision, you can show that the % loss of the initial KE depends only on the masses, not on the characteristics of the materials of the objects.

If the block that receives the bullet is "soft" then the friction force, f, on the bullet is relatively small and the bullet travels a good distance, d, into the block. For a hard block, f is greater but d is smaller. The work "done by friction", fd, is the same.

3. Dec 7, 2012

### mesa

Very good. So it seems the shape would not matter as well. Such a nice relation between energy, momentum, and Newtons Laws, it's quite wonderful.