Question

A puck with a mass m1 = 28.0 g moving at 1.00 m/s approaches a stationary puck with a mass m2 = 102 g on an air table and they undergo a two-dimensional elastic collision. As a result of their interaction, the incident puck moves away with a speed v1 = 0.785 m/s and the other puck moves away with a speed v2 in a different direction. What is the angle between the velocities v1 and v2 ?

Explain please!!!

Answer 1

m1 = 28.0 g
v1i = 1.0 m/s
m2 = 102.0 g
v2i = 0
v1f = 0.785 m/s
v2f = ?

Because Collision is Elastic, Initial KE = Final KE
1/2 * m1 * v1i^2 = 1/2 * m1 * v1f^2 + 1/2 * m2 * v2f^2
1/2 * 28 * 1.0^2 = 1/2 * 28 * 0.785^2 + 1/2 * 102 * v2f^2
v2f = 0.325 m/s

Let m1 and m2 diverge at angle θ & α Respectively.

Conservation of Momentum:
In the x direction
m1*v1i = m1*v1f*cos(θ) + m2*v2f*cos(α)
28*1.0 =  28*0.785*cos(θ) + 102*0.325*cos(α)
28 =  21.98*cos(θ) + 33.15*cos(α)
21.98*cos(θ) = 33.15*cos(α) - 28
(21.98*cos(θ))^2 = (33.15*cos(α) - 28 )^2   --------1

In the y direction
0 = m1*v1f*sin(θ) - m2*v2f*sin(α)
0 = 28*0.785*sin(θ) - 102*0.325*sin(α)
(21.98*sin(θ))^2 = (33.15*sin(α))^2 ----------2

Adding 1 & 2
(21.98*cos(θ))^2 + (21.98*sin(θ))^2 = (33.15*cos(α) - 28 )^2 + (33.15*sin(α))^2
21.98^2 * [(cos(θ))^2 + (sin(θ))^2 ] = 33.15^2 * [(cos(α))^2 + (sin(α))^2 ] + 28^2 - 2*33.15*28*cos(α)
21.98^2 = 33.15^2 + 28^2 - 1856.4 * cos(α)
α = 41.06

0 = 28*0.785*sin(θ) - 102*0.325*sin(41.06)
θ = sin^-1(21.77/21.98)
θ = 82.07

Angle between the Velocities = θ + α = 82.07 + 41.06
Angle between the Velocities = 123.13^o