A block of mass m = 2.02 kg slides down a 30.8 ° incline which is h = 3.46 m high. At the bottom, it strikes a block of mass M = 6.90 kg which is at rest on a horizontal surface. (Assume a smooth transition at the bottom of the incline.) If the collision is perfectly elastic, and friction can be ignored, determine the speed of the smaller block and larger box after the collision.
How would you determine the answer?
2006-12-03 10:20:56 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
Do you know Amanda C? This is the answer I gave her on this one.
You do have enough info. You can get the speed at the bottom by considering that the potential energy (mgh)will all be converted to kinetic ((1/2)mv^2) at the bottom of the ramp.
Then in a collision that is perfectly elastic, both momentum and energy are conserved. A bunch of math and the fact that the 2nd block is at rest gives the equations
v1 = (m1-m2)u1/(m1+m2)
v2 = 2m1*u1/(m1+m2)
2006-12-03 14:17:38 · answer #1 · answered by sojsail 7 · 0⤊ 0⤋