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Elastic Collision: ball held at angle released to hit a block attached to spring

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rj3214
#1
Nov12-12, 02:00 PM
P: 4
1. The problem statement, all variables and given/known data
A 200 g rubber ball is tied to a 1.0 m long string and released from rest at angle θ. It swings down at the very bottom has a perfectly elastic collision with a 1.0 kg block. The block is resting on a frictionless surface and is connected to a horizontal 20 cm long spring of spring constant 2000 N/m. After the collision, the spring compresses a maximum distance of 1.7 cm. From what angle was the rubber ball released?


2. Relevant equations



3. The attempt at a solution
I have no idea where to begin for this problem. Can somebody please point me in the right direction I'm not looking for the answer I just need to know how to start and I should be able to go from there
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Delphi51
#2
Nov12-12, 02:37 PM
HW Helper
P: 3,394
Everything is frictionless, which suggests an energy conservation approach. Do you have formulas for the various kinds of energy involved?
rj3214
#3
Nov12-12, 02:41 PM
P: 4
The only formula's I can think of to use are energy conservation:

1/2mvf^2 + 1/2kxf^2 = 1/2mvi^2 + 1/2 kxi^2

mghf + 1/2kxf^2 = mghi + 1/2kxi^2

Delphi51
#4
Nov12-12, 03:17 PM
HW Helper
P: 3,394
Elastic Collision: ball held at angle released to hit a block attached to spring

I think you will have to do it in three separate parts, first finding the speed of the ball just before it hits the block. Potential energy at top = kinetic energy at bottom.
The energy at the top will have the unknown angle in its expression, which you want to keep so you can solve for it in the end.

The second part is the collision. Likely the ball will bounce back, so some of the energy will stay with the ball while the rest is transferred to the block. What quantities are conserved in the collision?

The third part, compression of the spring, where the kinetic energy of the block is converted to spring energy should be easy. Maybe do parts 1 and 3 first, then tackle the more complicated part 2.
rj3214
#5
Nov12-12, 04:00 PM
P: 4
I understand where you're going but how can you find the velocity of the ball right before it hits the block. This is what I got for part 1: mgsinθ = 1/2mv^2
Delphi51
#6
Nov12-12, 06:15 PM
HW Helper
P: 3,394
That is the right equation - if θ is the angle from horizontal. It doesn't say, but I would have assumed angle with vertical and used cos θ instead of sin θ.
Cancel the m's and solve for v. You can't find a number for v without knowing θ but it should all work out after you do the other two parts.
rj3214
#7
Nov13-12, 07:48 AM
P: 4
Yeah you're right it is with respect to the vertical so I would have to use cosine. Can I just solve part 3 for velocity and then set that equal to what the velocity was in part 1 and solve for θ that way?
Delphi51
#8
Nov13-12, 11:05 AM
HW Helper
P: 3,394
No, you have to do the collision calc to connect the two motions. After the collision, the 1 kg block will not be moving at the same speed as the 0.2 kg ball was before.


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