A hard ball (1.0 kg) is moving 5 m/s and collides with an identical ball at rest. The collision istotally elastic, and in the center‐of‐mass frame the balls each deflect by 45 degrees. Find the velocities after the collision in the lab frame.
let m1 = m2 = m
let v1 and v2 are velocities of the two objects after the collision.
now apply conservation of momentum
0 = m1*v1*sin(45) - m2*v2*sin(45)
m2*v2*sin(45) = m1*v1*sin(45)
==> v1 = v2 (since m1 = m2)
now apply conservation of momentum in x-direction.
m1*u1 = m1*v1*cos(45) + m2*v2*cos(45)
u1 = v1*cos(45) + v2*cos(45)
5 = v1*cos(45) + v1*cos(45)
5 = 2*v1*cos(45)
v1 = 2.5/cos(45)
v1 = 2.5*sqrt(2)
= 3.54 m/s
so, v2 = 3.54 m/s
so, after the collsions both bodies move with velocities 3.54 m/s
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