Question

A 44g marble moving at 2.0m/s strikes a 20g marble at rest. Assume the collision is perfectly elastic and the marbles collide head-on. What is the speed of the first marble immediately after the collision? What is the speed of the second marble immediately after the collision?

Answer 1

let speed of first marbel be v1 in opposite direction to its initial direction and speed of second marbel be v2 in the direction of first marbel.

then conserving momentum,

44*2=20*v2-44*v1

-11*v1+5*v2=22..(1)

conserving energy

0.5*44*2^2=0.5*44*v1^2+0.5*20*v2^2

solving both the equations

we get

v1=-0.75 m/s

so first marbel moves in the same direction. with speed of 0.75 m/s

v2=2.75 m/s