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Conservation of Momentum problem

  • Thread starter Sheneron
  • Start date
360
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[SOLVED] Conservation of Momentum problem

1. Homework Statement

A 12.5 g wad of sticky clay is hurled horizontally at a 95 g wooden block initially at rest on a horizontal surface. The clay sticks to the block. After impact, the block slides 7.50 m before coming to rest. If the coefficient of friction between block and surface is 0.650, what was the speed of the clay immediately before impact?



2. Homework Equations
Ki = Kf + loss
Pi = Pf

3. The Attempt at a Solution

To solve this problem here is what I did:
Ki = Kf + loss
m1vi = (m1 + m2)vf

Using equation 2, I solved for vf. I then plugged that into the vf in the Kf in equation 1. Using this method, however, I did not get a correct answer. I feel like I am missing something at the beginning. Can someone tell me if this is the proper set up or if I am missing something? Thanks.
 

Answers and Replies

Doc Al
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Separate the problem into two parts:
(1) The collision itself. Use your momentum conservation equation here, just like you did.
(2) Post collision. What's the KE of the block+clay immediately after the collision? What's the work done by the friction force?
 
360
0
Would my 2 equations not be this:

1) (m1)(Vbefore) = (m1 + m2)(Vafter)

2) .5m1(Vbefore)^2 = .5(m1 + m2)(Vafter)^2 + loss
 
Doc Al
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44,827
1,083
360
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Ok, that worked.

Im not sure I completely understand why KE(after) = work done by friction.
 
Doc Al
Mentor
44,827
1,083
Because the work done on the block+clay equals the change in its KE. After the collision, it starts out with KE(after) and ends up with 0 as it comes to rest.
 
360
0
Simple enough. Thanks
 

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