How Do You Solve a Collision Problem with a Pendulum to Maximize Height?

[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 .

 

[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?

 

[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)MV^{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?

 

[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.

 

[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.

 

[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.)

 

[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 !

 

[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.