Question

A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 5.05 m/s at an angle of 32.7° with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision.

magnitude	_____________ m/s
direction	____________ ° counter-clockwise from the original direction of motion

Answer 1

Given,

A billiard ball moving at 6.00 m/s

first ball moves at 5.05 m/s

angle is 32.7°

let m = mass of each

so, monent in Y drection :

m x 5.05 x sin 32.7 = m x v2 x sin (t)

after collison speed 2nd ball = v2

2.728 = v2 x sin t ------(1)

moment in X -direction :

m x 6= m x 5.05 x cos 32.7 + mx v2 x cos t

so, v2 x cos t = 1.75-------- (2)

so (1) / (2) we get :

tan (t) = 1.558

angle = t = tan^-1 (1.558)

= 57.30 degree

v2 = 1.558/ (sin 57.30)

  v2 = 1.85 m/s

angle 57.30 degree