Question

Billiard ball A of mass mA = 0.122 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.145 kg . As a result of the collision, ball A is deflected off at an angle of θ′A = 30.0∘ with a speed v′A = 2.10 m/s, and ball B moves with a speed v′B at an angle of θ′B to original direction of motion of ball A.

1) Solve these equations for the angle, θ′B, of ball B after the collision. Do not assume the collision is elastic.

2)Solve these equations for the speed, v′B, of ball B after the collision. Do not assume the collision is elastic.

Answer 1

use the law of conservation of momentum as

mA vA - mA vA' - mB vB' = 0

0.122 * (2.8) [1,0] - 0.122 * (2.10)*(cos 30, sin 30) - vB' (0.145) (cos (theta), sin (theta)) = 0

sUBSTITUING And calculating properly

vB' (0.14) cos(theta = 0.122 * (2.8) - 0.122 (2.10) cos 30 ----------------- 1

vB' (0.14) sin (theta) = 0 - 0.122 (2.10) sin 30. ---------------------------------------2

equation 2 divided by equation 1,

SOLVING FOR ANGLE THETA

WE GET

tan (theta) = (0 - 0.122 (2.10) sin 30)/(0.122 (2.80) - 0.122 * (2.10) * cos 30)

tan (theta) = -0.1281/0.1198

tan (theta) = -1.069282

(theta) = -46.91 degree--------------<<<<<<<<<<<<<<<Answer to part A

---------------------------------------------------------------------

AGAIN CHECKING BACK FROM EQN 2 ,

0 - 0.122 (2.10) sin 30 = vB' (0.140) sin (-46.91�)

-0.1281 = vB' * - 0.102239

vB' = 1.253 m/s----------><<<<<<<<<<<<<<<<<<Answer to part B