How Do You Calculate Velocity and Energy in an Elastic Collision?

AI Thread Summary
In an elastic collision between blocks A (3.50 kg) and B (10.00 kg), block A moves at 9.00 m/s while block B is initially at rest. The maximum energy stored in the spring bumpers is determined by the difference in kinetic energy before and after the collision. The key to solving for the velocities of both blocks at maximum compression is recognizing that their relative velocities are zero at that point. Using conservation of momentum, the final velocities of both blocks are calculated to be 2.33 m/s, with the elastic potential energy reaching 105 J. This approach effectively combines kinetic energy and momentum principles to solve the problem.
irun4edmund
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Homework Statement


Blocks A (mass 3.50 kg) and B (mass 10.00 kg) move on a frictionless, horizontal surface. Initially, block B is at rest and block A is moving toward it at 9.00 m/s. The blocks are equipped with ideal spring bumpers. The collision is head-on, so all motion before and after the collision is along a straight line. Let +x be the direction of the initial motion of A.

Find the maximum energy stored in the spring bumpers and the velocity of each block at the time of the collision.


Homework Equations


K = 0.5mv2
Wnet = Kf - Ki

The Attempt at a Solution


Since the block A is the only block in motion wouldn't the elastic potential energy be equal to the kinetic energy? I tired using 0.5m(v)^2 and setting that equal to the Kinetic energy, but i didn't get the right answer (i got 141.75).

As far as solving for the velocity of blocks A and B, I'm not sure how to go about it. i tried using Vaf = [(3.5-10.0) / 13.5] * 9.00 = 4.3 but this is the velocity of block A right after the collision not during (answer to another part of the question, but it doesn't help me with the first part).

Any help on this would be greatly appreciated. thanks guys.
 
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irun4edmund said:
Find the maximum energy stored in the spring bumpers and the velocity of each block at the time of the collision.

Since the block A is the only block in motion wouldn't the elastic potential energy be equal to the kinetic energy?

Hi irun4edmund! Welcome to PF! :smile:

No, elastic PE = KE before minus KE after.

Hint for velocities: when the spring is at maximum compression, what is the relative velocity of the blocks?
 
ok i tried PE = 0.5mavai2 -0.5mavaf2-0.5mbvbf2 and got 3.55 J. That wasn't right either...

If the compression of the springs was at an maximum, wouldn't the relative velocities be zero? that was my intial guess and that wasn't right either.
 
Last edited:
irun4edmund said:
If the compression of the springs was at an maximum, wouldn't the relative velocities be zero? that was my intial guess and that wasn't right either.

Hi irun4edmund! :smile:

That is correct … then you combine vaf = vbf with conservation of momentum to find what they are. :smile:
 
Oh.. my.. god. It worked. I never would have thought to set Va = Vb I though i was working 2 equations with 3 unknowns.

I got Va = Vb = 2.33 m/s

Elastic Potential = 105 J

Thank you!
 
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