# Thread: Kinetic energy lost to an inelastic collision, Lab question

1. ## Kinetic energy lost to an inelastic collision, Lab question

We did the ballistic pendulum lab. Variables are defined as in the diagram. Additionally, let h = the height change in the block from rest to when it swings to its highest point.

Our goal was to determine v by measuring the average angle $\theta$ over 10 trials. Here's the question that's got me stuck:

Is the kinetic energy before and after the collision conserved? If not, where did the energy go/come from?

What I did:

$m=.0077kg, M=.0804kg, (m+M)=.0881kg, \theta=14^{\circ}, L=.0227m$

$PE_1+KE_1=PE_2+KE_2$ for the system after the collision

$KE_1=PE_2$ assuming the block starts at height of 0

$\frac{1}{2}(m+M)(v')^2=(m+M)gh$

$v'=\sqrt{2gh}$

Now we look at the whole system using conservation of momentum:

$mv=(m+M)v'$

$v=\frac{(m+M)\sqrt{2gh}}{m}$ solving for v and substituting for v'

$h=L(1-cos(\theta))$ Some trig to get h in terms of theta

$v=\frac{(m+M)\sqrt{2gL(1-cos(\theta))}}{m}$ Substitute for h

This is all just to show where I got v and v', and how it is all directly from our measured angle of theta, the measured masses of the balls, and length of the pendulum. Now I find if the kinetic energy stayed the same:

$KE_1=\frac{1}{2}mv^2$

$KE_1=\frac{1}{2}m\left(\frac{(m+M)\sqrt{2gL(1-cos(\theta))}}{m}\right)^2$

$KE_1=\frac{(m+M)^2gL(1-cos(\theta)))}{m}$

I plugged in all of the numbers and got:

$KE_1=.00666J$

$KE_2=\frac{1}{2}(m+M)(v')^2$

$KE_2=\frac{1}{2}(m+M)(\sqrt{2gL(1-cos(\theta)})^2$

Plug in the numbers again:

$KE_2=.000583$

So they differ by a factor of about 11. I can't figure out exactly why. My first guess would be that energy is lost due to the ball heating up the block, the block being given some rotational momentum, or maybe friction in the part of the apparatus that measures the angle. The problem with these answers is that it seems like the measurements shouldn't be affecting this difference in energies. That is, I found both v and v' as functions of measured values that didn't change before and after the collision, so it seems like I can't say the difference in kinetic energy I calculated is due to experimental error. So if that's the case, where's that energy at?

2. ## Re: Kinetic energy lost to an inelastic collision, Lab question

All your working is correct. Some energy is converted to sound. If no energy was lost you would have the equation 1/2mv^2=1/2(m+M)v'^2
In collisions momentum is conserved and energy lost.