F = MA Exam 2010 #'s 15,16 (Two mass system with initial velocity)

[SignaturePF]

Homework Statement

See:
http://www.aapt.org/physicsteam/2010/upload/2010_FmaSolutions.pdf
#15, 16

Homework Equations

U_o + K_o = U + K
v_CoM = vM1 + vM2 / (M1 + M2)

The Attempt at a Solution

I thought that for 15 it was simply:
1/2mv^2 = mgh
h = v^2/2g

But this is wrong, so then I looked at the answers and saw that they dealt with CoM. How am I supposed to do this problem? And what should have hinted me toward using the CoM besides the answer choices?

 

[tiny-tim]

Hi SignaturePF!

(try using the X² button just above the Reply box )

I thought that for 15 it was simply:
1/2mv^2 = mgh
h = v^2/2g

ah, but you're assuming the KE at the top is 0, and it isn't

common-sense (if nothing else! ) tells you that at the top, the velocities of both blocks must be the same

(and as a general rule, conservation of momentum will almost always need to be fitted in somewhere)

show us what you get ​

 

[ModusPwnd]

When I read the problem I first thought "Conservation of Energy!". Why? Because its easier than kinematic equations and is often the key to a problem. But upon further inspection I quickly came into a issue... The triangular mass is not fixed to the table and is free to slide... This means that I can't equate the initial kinetic energy of 'm' to the final potential energy of 'm'. I have to also consider the final kinetic energy of 'M'. When I write that equation out I have too many variables and can't solve it. To eliminate another variable I appeal to another equation, conservation of momentum. This is the natural thing to do when solving problems like this. Conservation of momentum should be thought of just after (or maybe even before) conservation of energy. Particularly in these types of problems. Go into the problem knowing that you may need to pull one, or both out. Also recall generally if you have one unknown you can solve it with one equation but for two unknowns you need two equations. If you have too many variables and and want to eliminate one (or more) you should think to use conservation of energy or momentum (along with any other pertinent equations of course).

 

[SignaturePF]

Ok, let's see.
Obviously, as both of you mentioned, it's a perfectly inelastic collision; that is, the two velocities are the same.

mv = (m+M)v'
v' = m/(m+M)v

Now we apply conservation of energy:
U + K = U' + K'
U initial is zero
For K, the triangle and block both start off with v' and it ends up with just the triangle moving while the block eventually stops right?

So:
1/2m²/(m+M)²v² = mgh + 1/2Mv²
Subtracting right side KE from both sides
(m^2/(M+m) - M)v^2 = 2mgh
Ya this is messed up.

 

[tiny-tim]

v' = m/(m+M)v

yes

but you haven't applied it correctly here …

1/2m²/(m+M)²v² = mgh + 1/2Mv²

(and shouldn't the last term should be 1/2mv² ?)

 

[SignaturePF]

Ok let me try again. Is the energy thought process correct though.
1/2(M+m)((m/(M+m)v)^2 = mgh + 1/2Mv^2
I think the last term should not be mv^2 because we are looking for when the block has STOPPED. It should be Mv^2 because I thought that the triangle/plane would continue moving.
But simplifying using what you said:
1/2m^2/(M+m)v^2 = mgh + 1/2mv^2
Multiply by 2, divide by m
m/(M+m)v^2 = 2gh + v^2
Subtract v^2:
v^2(m/(M+m) - 1 ) = 2gh
v^2(-M /(M+m) = 2gh - Why is there a negative sign?

h = v^2(M/M+m)/2g
This is the correct answer. But I'm still a bit confused.

 

[tiny-tim]

I think the last term should not be mv^2 because we are looking for when the block has STOPPED. It should be Mv^2 because I thought that the triangle/plane would continue moving.

your RHS should be the initial KE, which is for the small moving mass, m

Why is there a negative sign?

the final KE must be less than the initial KE,

so yes, you must subtract mgh from the initial KE

 

[SignaturePF]

I don't think I understand what you're saying. Could you rephrase that?

 

[tiny-tim]

the initial KE is all you've got

everything else has to come from that

so initial KE = final KE + PE ​

(and now I'm off to bed :zzz:)