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Block of mass sliding down incline and colliding

  • Thread starter AnnieF
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  • #1
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Homework Statement



A block of mass m=2.2 kg slides down a 30 degree incline which is 3.6 m high. At the bottom, it strikes a block of mass M=7 kg which is at rest on a horizontal surface. If the collision is elastic, and friction can be ignored, determine (a) the speeds of the two blocks after the collision, (b) how far back up the incline the smaller mass will go, and (c) what is the upper limit on mass m if it is to rebound from M, slide up the incline, stop, slide down the incline, and collide with M again?

Homework Equations



mgh=1/2mv^2
mghsintheta=1/2mv^2
mv1+mv2=mv1'+mv2'


The Attempt at a Solution



a) mgh=1/2mv^2
v=8.4m/s
v'a=(2.2-7/2.2+7)(8.4)=-4.38 m/s
v'b=(2*2.2/7+2.2)(8.4)= 4.02 m/s

b) mghsintheta=1/2mv^2
h=1.95 m

I am not sure if I did a and b right and I have no idea how to set c up. Any help would be appreciated!
 

Answers and Replies

  • #2
cepheid
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Science Advisor
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Welcome to PF AnnieF!

Your equations and your arithmetic look good for part (a).

For part (b):

I agree that the force acting on the mass in the direction parallel to the ramp is given by mgsin(θ). However, this force has to be multiplied by the displacement of the mass in order to get the work done on the mass, which should be its change in KE. The displacement is not equal to h, so your equation is not correct.

Another way to think about it: conservation of energy says that if the mass goes up the ramp and stops, all of the mass's initial KE should be converted to potential energy, which would give us the formula:

KE = mgh

which isn't consistent with the equation you used.

For part (c):

You know that the velocity after the collision is given by:

va = [ (m-M)/(m+M) ]*v

Since there is no friction, the kinetic energy that the mass m will have after sliding up the incline, stopping, and sliding back down again will be equal to the kinetic energy it had just after the collision. (Make sure you understand why this is true, and if you don't, ask me about it!). Therefore, its speed will be the same as it was directly after the collision, only now it will be travelling towards mass M instead of away from it. So, what condition on va must be satisfied if mass m is to eventually collide with mass M, given that it is travelling in the same direction?
 
  • #3
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Thank you for the correction on part b. I understand the mistake. On part C, however, I am still slightly confused. I understand the part about the kinetic energy being the same after sliding up, stopping and sliding back down as it was after the collision, but I don't understand the second part. If the velocity is the same, then shouldn't the momentum also be the same? I tried using momentum conservation, but there are too many unknowns and I am unsure of how to get the upper limit on the mass.
 
  • #4
cepheid
Staff Emeritus
Science Advisor
Gold Member
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35


I understand the part about the kinetic energy being the same after sliding up, stopping and sliding back down as it was after the collision, but I don't understand the second part. If the velocity is the same, then shouldn't the momentum also be the same?
The momentum would not be the same -- its magnitude would be the same, but it would have the opposite direction.

I tried using momentum conservation, but there are too many unknowns and I am unsure of how to get the upper limit on the mass.
Momentum conservation is not relevant here. You're not trying to figure out what happens after the second collision. You're trying to figure out what must be true in order to make sure that collision happens in the first place. The key to doing that is answering this question:

what condition on va must be satisfied if mass m is to eventually collide with mass M, given that it is travelling in the same direction?
Can you answer this question? Hint, in order for mass m to collide with mass M on the way back, it has to catch up to it first, doesn't it?
 

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