# Velocity - conservation of momentum and kinetical energy

1. Jan 12, 2012

### mstud

1. The problem statement, all variables and given/known data

The speed of a projectile is measured by shooting it into a sanbox hanging in 6.00 m long threads. The projectile has a mass of 12.0 g, and the sandbox has a mass of 3.00 kg. The projectile ends up laying still in the sandbox. Thus the sandbox and the projectile moves as one body after the collision. The pendulum with the sandbox swings out and up after the shot until it reaches it's largest angle with the vertical line, 22.5 degrees.

a) Find the speed of the projectile before it hits the sandbox.

b) How much kinetical energy is lost during the collision?

Answers:

a) 751 m/s, b) 3.37 kJ.

2. Relevant equations

Linear momentum: p=mv

Conservation of linear momentum: $p_{before}=p_{after} \Leftrightarrow m \cdot v_0=(m+M)v$

where m is the mass of the projectile and M the mass of the sandbox, v_0 the initial velocity of the projectile and v the velocity of the projectile-sandbox body.

Conservation of mechanical energy: $mgh=\frac 12 mv^2$ where m is mass, g is 9.81 m/s^2, h is height and v is velocity.

3. The attempt at a solution

a)

I tried to calculate the speed of projectile-sandbox just after the collision by using that it's swinging up with mechanical energy being conserved, but it seems I don't get the right answer (mechanical energy is conserved but kinetical energy is not, right?) :

I must use trigonometry to find the height the sandbox swings up.

Height from suspension to sandbox: $\frac {x}{\sin 67.5}=\frac {6.00}{\sin 90} \Leftrightarrow x= 6.00 \cdot \sin 67.5 = 5.54 m$. This means the sandbox has swung 6.00 m - 5.54 m = 0.46 m upwards.

In this highest point the velocity of the body is 0, and its potential energy is $mgh= (0.012 kg + 3.00 kg) \cdot 9.81 m/s^2 \cdot 0.46 m=13.6 J$.

This means the body must have had an initial velocity of $\frac 12 mv^2= 13.6 J \Leftrightarrow v=\sqrt{2\cdot 13.6}=5.215 m/s$.

The initial velocity of the projectile is then:

$m_{projectile} \cdot v_{projectile} = m_{body} \cdot v_{body} \Leftrightarrow v_{projectile}=\frac{m_{body} \cdot v_{body}}{m_{projectile}} =\frac{(0.012 kg+3.00 kg)\cdot 5.215 m/s}{0.012 kg}=1.309 \cdot 10^3 m/s$, which is not the correct answer.

Can anybody please help me get this right?

Last edited by a moderator: Jan 12, 2012
2. Jan 12, 2012

### technician

your method is perfectly correct but you have miscalculated the velocity of the box after the collision
you have 1/2(mv^2) = 13.6 J and I think you have forgotten to include 'm' in your calculation.
I got v = 3 m/s

3. Jan 12, 2012

### Curious3141

Mistake here. You forgot to divide by the mass (which should actually be called (M+m) as per your nomenclature).

Much easier to manipulate this symbolically:

$\frac{1}{2}(M+m)v^2 = (M+m)gh = (M+m)gL(1 - \cos\theta)$ where $\theta$ is the angle subtended from the vertical, as given.

Then $v = \sqrt{2gL(1 - \cos\theta)}$ which is much less error-prone.

A word of advice: whenever doing multi-part mechanics (or even physics in general) problems, try to manipulate everything symbolically until the very last stage when you plug in the values. The benefit is that this process is less likely to cause errors like this, with you being able to keep track of what's going on (conceptually) at every stage; you're also able to reduce intermediate round-off errors. And in some problems, it's absolutely essential you learn how to do this as you need things to cancel out, otherwise you just can't solve the problem (quite typical of Olympiad type mechanics problems).

And please keep your variables consistent.

Last edited: Jan 12, 2012
4. Jan 12, 2012

### mstud

Thanks a lot for your correction and advice!

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