- #1

- 29

- 0

Consider a frictionless track as shown in Figure P6.48. A block of mass m1 = 5.55 kg is released from A. It makes a head on elastic collision at B with a block of mass m2 = 11.0 kg that is initially at rest. Calculate the maximum height to which m1 rises after the collision.

I used to .5mv^2 = mgh to get the velocity of the first block at the bottom. That was 9.9 m/s.

Then I did m1vi + m2vi = m1vf + m2vf => m1vi + 0 = 0 + m2vf to solve for the speed of the second block when the first block has gone back up the ramp and stopped. I then used the energy in the first equation and conservation of energy to solve for the final height.

KE = m1gh + .5m2v2f

But this gave me a height of 4.49 meters which WebAssign says is not correct. Thanks for any help. By the way, it's due at 8:30 EST tomorrow morning.

I used to .5mv^2 = mgh to get the velocity of the first block at the bottom. That was 9.9 m/s.

Then I did m1vi + m2vi = m1vf + m2vf => m1vi + 0 = 0 + m2vf to solve for the speed of the second block when the first block has gone back up the ramp and stopped. I then used the energy in the first equation and conservation of energy to solve for the final height.

KE = m1gh + .5m2v2f

But this gave me a height of 4.49 meters which WebAssign says is not correct. Thanks for any help. By the way, it's due at 8:30 EST tomorrow morning.

Last edited: