# Basic Collision/Pendulum Problem

1. Nov 26, 2007

### donwa83

Ok, I need help with a basic collision problem. Please see the attached image and picture the image as a pendulum.

You bring mass 'M' up a certain height and let it go. It collides elastically with mass 'Mew M." Mass 'm' goes up a maximum height. Find mass 'Mew M' so that mass 'm' can attain the maximum height.

Also:
(1) M Does not equal m.
(2) The answer is not infinitely small, which is what i got .

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2. Nov 26, 2007

### Shooting Star

Why don't you show us how you got the wrong answer? Maybe we can work it out from there.

What things remain conserved in elastic collisions?

3. Nov 26, 2007

### donwa83

Energy is conserved in a completely elastic collision. I used (1) energy of 'M', (2) collision of 'M' and 'm', (3), collision of 'Mew' and 'm', (4) energy of 'm'

First see attached for description of initial height.

So:

(1) Energy
Mg(L-Lcos$$\theta$$ )=(1/2)M$$V^{2}$$

V = $$\sqrt{2gL(1-cos\theta$$)

(2) Find the Velocity of 'Mew' after the collision, $$V_{3}$$, with momentum equation.

$$V_{3} = \frac{M(V-V_{2})}{Mew}$$

$$V_{2}$$ = Velocity of M after collision with Mew

(3) Find the velocity of 'm' after the collision with 'Mew,' $$V_{5}$$ with momentum equations.

$$V_{5} = \frac{M(V-V_{2}) - MewV_{4}}{m}$$

$$V_{4}$$ = the velocity of 'm' after the collision with 'Mew'

(4) Use energy equation to find Hmax.

$$mgHmax = \frac{1}{2} m V_{5}^{2}$$

Am I on the right track? Or am I missing something?

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4. Nov 27, 2007

### Shooting Star

You need not worry about the height. The only thing of concern is the KE and V of the mass M at the moment of collision. The values of the KE etc is immaterial, because you have to find mew-m in terms of M and m.

When a mass M collides with another mass, say M', what is the value of M' such that the whole KE of M is transferred to M' ? Think on these lines.

5. Nov 27, 2007

### donwa83

I got it?

When a mass M collides with another mass, say M', what is the value of M' such that the whole KE of M is transferred to M' ? Think on these lines.

M transfers all its kinetic energy to M' only if M' is the same mass as M!! So the answer is Mew-M's mass must be equivalent to the M's mass. Or will 'm' go further if Mew-M was equal to m and not M? M does not equal m.

6. Nov 27, 2007

### Shooting Star

Not mew_m' mass but the combined mass of mew_m and m!

Let's assume M comes to rest and gives up all the KE to mew-m and m. Then mew_m and m must add up to M.

You may ask whether that'll ensure the mass m receiving the max energy. You see, not only the KE but the momentum MV of the mass M is also being transferred to mew_m+m. Both conditions combined make m recv the max KE, and so swing up the highest.

(The delay in my first reply was due to a messy piece of algebra I had to before I was satisfied that this was indeed the case.)

Last edited: Nov 27, 2007
7. Nov 27, 2007

### donwa83

I understand your answer conceptually, but I'm not sure how to solve it algebraically. Should I start with a collision between M and Mew (which will yield the same velocity) and then Mew with m?

Also, initially I was thinking that m must be:

m = $$\frac{1}{2}$$ M + $$\frac{1}{2}$$ Mew

Ah, its so open ended without math !

8. Nov 27, 2007

### Shooting Star

Let MV = p and (1/2)MV^2 = k, for easier wrtining. I'll call mew_m as m1 and its speed as v1 etc.

m1v1^2 + mv^2 = 2k and m1v1 + mv = p. k and p are constants.

Eliminate v1 from the two eqns and maximise v. We want v to be max because m is a constant in the problem, so (1/2)mv^2 will be max.

That'll give you the reqd condition.