1. Oct 13, 2009

stella77

1. The problem statement, all variables and given/known data

A 200 g block is released from rest at a height of 25 cm on a frictionless 30 degree incline. it slides down the incline and then along a frictionless surface until it collides elastically wt an 800 g block at rest 1.4m from the bottom of the incline. how much later do the 2 blocks collide again?

mgh = 0.5mv^2
0.2 x 0.98 x 0.25 = 0.5 x 0.8 x v^2
0.49 = 0.4 v^2
v=1.11 m/s

and what should I do after this??

thx for the help!!!

2. Relevant equations

3. The attempt at a solution

2. Oct 13, 2009

dotman

Hello,

This is a bit of a tricky problem. Interesting though. Have you drawn a picture of whats happening here? If you haven't, you definitely should. I see you calculated the velocity of the first block as it reaches the bottom of the ramp; I didn't check your math, but the method looks correct.

So what's going to happen? That block is going to slide down the ramp, across the surface, and impact the second block. When it does, its going to impart some of its energy to that second block, putting that block into motion. At the same time, the first block will bounce off, backwards, and head back up the ramp (since there is no friction). Eventually it will come to a stop, slide back down the ramp, and begin heading again towards the second block, which has been sliding away all this time. If it has sufficient speed, it will overtake that block.

So first you need to find out what the speeds of each block will be after the collision, and once you have that, you can incrementally calculate what time they will recollide. What equations do you have to deal with an elastic collision? Remember that in an elastic collision, both momentum and kinetic energy are conserved.

Hope this helps. This problem is a bit complicated-- please reply with further questions!

3. Oct 13, 2009