Perfectly elastic collision?


by mattpd1
Tags: collision, elastic, perfectly
mattpd1
mattpd1 is offline
#1
Oct17-10, 02:08 AM
P: 13
1. The problem statement, all variables and given/known data
A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0m high frictionless chute to a waiting truck. Unfortunately, the truck driver went on a break without having removed the previous package, of mass 2m, from the bottom of the chute.

Suppose the collision between the packages is perfectly elastic. To what height does the package of mass m rebound?

http://session.masteringphysics.com/...3/3/10.P42.jpg




2. Relevant equations




3. The attempt at a solution

So far I know that the velocity of mass m at the moment of impact is 2.6 m/s. I think I should be using the conservation of energy/momentum to solve for the final velocity of mass m. I then think I can use this velocity to solve for K, and use this to solve for the height from U=mgh. I just can't figure out how to solve the final velocities.

[tex]2.6m=mv_{1}_{f}+2mv_{2}_{f}[/tex]

[tex]\frac{1}{2}m(2.6)^2=\frac{1}{2}mv_{1}_{f}^2+mv_{2}_{f}^2[/tex]

Are these set up right? If so, how do I continue? As you can see my algebra skills may be lacking.
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Quinzio
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#2
Oct17-10, 03:45 AM
P: 558
First of all I see that "m" appears everywhere. Can you simplify it ?
willem2
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#3
Oct17-10, 03:49 AM
P: 1,351
Your result of 2.6 m/s is wrong. You can calculate this with (final kinetic energy) = (initial potential energy), so (1/2)mv^2 = mgh

Apart from this your equations are correct. For a start you can divide them by m.

There are 2 ways of doing this. The first is to rearrange the first equation to get v_1 as a function of v_2, and then substitute this in the second equation.

A better way is to find the velocity of the center of mass, and compute all velocities in the reference frame of the center of mass. The equations become

[tex] 0 = mv_{1}_{f}+2mv_{2}_{f} [/tex]

(0 because it's the center of mass frame)

[tex] KE = \frac{1}{2}mv_{1}_{f}^2+mv_{2} _{f}^2 [/tex]

where KE is the initial kinetic energy in this frame

It's then easy to see that if (v1,v2) is a solution to these equations, so is (-v1,-v2)

Finally you have to transform the velocities back to the rest frame.

mattpd1
mattpd1 is offline
#4
Oct17-10, 10:42 PM
P: 13

Perfectly elastic collision?


Quote Quote by willem2 View Post
Your result of 2.6 m/s is wrong. You can calculate this with (final kinetic energy) = (initial potential energy), so (1/2)mv^2 = mgh

Apart from this your equations are correct. For a start you can divide them by m.

There are 2 ways of doing this. The first is to rearrange the first equation to get v_1 as a function of v_2, and then substitute this in the second equation.

A better way is to find the velocity of the center of mass, and compute all velocities in the reference frame of the center of mass. The equations become

[tex] 0 = mv_{1}_{f}+2mv_{2}_{f} [/tex]

(0 because it's the center of mass frame)

[tex] KE = \frac{1}{2}mv_{1}_{f}^2+mv_{2} _{f}^2 [/tex]

where KE is the initial kinetic energy in this frame

It's then easy to see that if (v1,v2) is a solution to these equations, so is (-v1,-v2)

Finally you have to transform the velocities back to the rest frame.
Oh crap. 2.6 is the speed of the objects after an inelastic collisions. I have been using the wrong number the whole time. Let me try it the right way... Is the velocity 7.67 m/s?

GOT IT! 33cm is the rebound height.

When I did finally get the correct FINAL velocity for the mass, it was -2.6. Is this just a coincidence that this is the same as the 2.6 that I thought was the initial velocity? Anyway, thank you!!!


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