What is the final height reached by block 1 after a perfectly elastic collision?

In summary, two blocks of masses 5.00 kg and 10.0 kg slide along a frictionless track. The first block is released from point A and the second block sits at the bottom of a ramp. The blocks collide at position B in a perfectly elastic collision. The problem asks for the height that the first block will rise after the collision. To solve this, the conservation of momentum and energy must be used to determine the speed of the first block before and after the collision. The final speed can then be used to calculate the height it will rise.
  • #1
bearhug
79
0
2 blocks are free to slide along the frictionless wooden track. The block of mass m1=5.00 kg is released from A, while the block of mass m2= 10.0 kg initially sits @bottom of ramp. The blocks collide @ position Bin a perfectly elastic collision. To what height does m1 rise after collision?

Originally I thought of using Ki + Ui = Kf + Uf where U=mgh
so 1/2m1vi^2 1/2m2vi^2 + (mgh)i = 1/2m1vf^2 1/2m2vf^2+ (mgh)f
However I'm having a hard time figuring this out because I don't know what the velocities of either block is after collision. I do know that the initial of block 2 is 0 m/s. Can someone help me set this problem up? Any help is appreciated.
 
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  • #2
What else is conserved in every collision?
 
  • #3
In every collision... momentum
 
  • #4
In elastic collisios kinetic energy
 
  • #5
bearhug said:
In every collision... momentum
Right. You have to use that fact to solve this problem. (You just used conservation of energy--but that's not enough.)
 
  • #6
Does gravitational potential energy have anything to do with this problem?
 
  • #7
OK so I set up the problem beginning with (m1v1 + m2v2)i = (m1v1 + m2v2)f . Should I start this problem with block 1 m=5.0 kg w/ initial velocity after collision or before. If it's before than initial would be 0 but other wise it wouldn't. Since the question asks for the height after collision I was wondering if this needs to be considered in terms of what's initial and what's final. Any feedback please.
 
  • #8
Assuming I understand the problem correctly, here's how to approach it. First figure out the speed of block 1 just before it collides with block 2. Then analyze the collision to determine the speed of block 1 just after the collision. (That involves conservation of momentum and energy.) Once you know the speed of block 1 after the collision, figure out how high it goes.
 

What is a perfectly elastic collision?

A perfectly elastic collision is a type of collision in which there is no loss of kinetic energy. This means that the total kinetic energy of the system before and after the collision remains the same.

What are the characteristics of a perfectly elastic collision?

In a perfectly elastic collision, the objects involved bounce off each other without any deformation or loss of kinetic energy. The total momentum and kinetic energy of the system is conserved.

What is the formula for calculating the final velocities in a perfectly elastic collision?

The formula for calculating the final velocities in a perfectly elastic collision is v1f = (m1 - m2) / (m1 + m2) * v1i and v2f = (2 * m1) / (m1 + m2) * v1i, where m1 and m2 are the masses of the objects and v1i is the initial velocity of the first object.

Can a real-life collision be perfectly elastic?

No, a perfectly elastic collision is an idealized concept and cannot occur in real-life situations. There will always be some loss of kinetic energy due to factors such as friction and deformation of the objects involved.

What are some examples of perfectly elastic collisions?

Some examples of perfectly elastic collisions include two billiard balls colliding, atoms colliding in a gas, and photons bouncing off a mirror.

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