Will the cart reach the end of the ramp?

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Homework Statement


A student flings a 23g ball of putty at a 225g cart sitting on a slanted air track that is 1.5m long .The track is slanted at an angle of 25° with the horizontal. If the putty is traveling at 4.2m/s when it hits the cart, will the cart reach the end of the track before it stops and slides back down?


Homework Equations


Conservation of Momentum, Energy
Work


The Attempt at a Solution


I set up the diagram so that the slant is the hypothenuse being 1.5m long at 25° to the horizontal, and found the opposite length (height) being 1.5sin25° = 0.634m

I then solved for how much potential energy the cart will have if it reaches the end of the ramp.
Ep = (0.0225)(9.81)(0.634)
= 1.586J

I am now confused as to how to get the velocities, because writing out the conservation of momentum equations leads to me running into two unknown variables; the speed of the putty after, and the speed of the cart after.

Would anyone mind helping me out further with this problem..
 
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I believe that the putty sticks to the cart. That should help.
 
Ok, wait WHAT is a putty..É -- sorry I canèt write question marks for some reason,,
 
Similar in consistency to bread dough, for instance.
 
Alright, so would it work out if I countinued on like this:
Let "c" subscript rep the cart
Let "p" subscript rep the putty

MPVP + 0 = (MP+MC)(V'PC)
MPVP
------ = (V'PC)
(MP+MC)

From this I get 0.3987m/s. Then I put this into the conservation of Energy scenario where:

Ek = Eg'
So LS must equal, or be close to RS if the cart were to reach the top
LS:
(1/2)(0.023kg + 0.225kg)(0.3987m/s)^2
= 0.0494J

RS:
(0.023kg + 0.225kg)(9.81m/s^2)(1.5sin25°)
= 1.542J
Therefore, the cart did NOT reach the top of the inclined surface..
Is what I am doing correct?
 
In fact, you could replace the 1.5 with x (or whatever variable you prefer) to find out how far up the ramp the cart (with putty) will go.

What you did looks fine.