Question

The white ball in the figure has a speed of 1.45 m/s and the yellow ball is at rest prior to an elastic glancing collision. After the collision the white ball has a speed of 1.430435 m/s. Measured counterclockwise from east, what angle does it scatter at, if the yellow ball is scattered at 280 degrees, also measured counterclockwise from the east?

Answer 1

for elastic collsion , Applying energy conservation,  

KEi = KEf

(2 x 1.45^2 /2) + 0 = (2 x 1.430435^2 / 2) + (1 v^2 / 2)

v = 0.336 m/s

Applying momentum conservation,

(2 x 1.45)i + (1 x 0) = 2 x 1.430435 (cos(theta)i + sin(theta)j) + (1 0.336 (cos280 i + sin280j))

2.90 i = (2.8608 cos(theta) + 0.0582)i + (2.8608 sin(theta) - 0.3309)j

so sin(theta) = 0.3309/2.8608  

theta = 6.64 deg

so 2.8608 cos(theta) + 0.0582 = 2.90

theta = 6.6 deg

Ans: 6.6 deg