A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. As a result, the incident ball is deflected 30.00 from its initial...

Question

A 2.5 g steel ball moving 0.40 m/s collides elastically with a stationary, identical ball. As a result, the incident ball is deflected 30.00 from its initial path.
a. What is the velocity of the incident ball after the collision? (ans: 0.35 m/s, 30° to incident direction)
b. What is the velocity of the struck ball after the collision? (ans: 0.20 m/s, 60° to incident ball direction)

Answer 1

Conserve kinetic energy and linear momentum: this generates THREE equations that let you solve for the unknowns.

m1 is the incident ball

m2 is the struck ball, note that is the same size as m1

v0 is the incident ball's initial speed = 0.4 m/s

v1 is the incident ball's final speed

v2 is the struck ball's final speed

theta1 is the incident ball's final angle = 30 degrees

theta2 is the struck ball's final angle

 

Momentum in each direction gives:
Horizontal: m1 v0 = m1 v1 cos theta1 + m2 v2 cos theta2

Vertical: 0 = m1 v1 sine theta1 - m2 v2 sine theta2

NOTE: m2 is deflected BELOW the path.

 

Kinetic energy:

1/2 m1 v0^2 = 1/2 m1 v1^2 + 1/2 m2 v2^2

 

NOTE: all masses are equal so they immediately cancel out in all equations.

 

Have fun with the quadratic equation!

 

bill