Question

Billiard ball A of mass mA = 0.125 kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140 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.

Part C

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

Express your answer to three significant figures and include the appropriate units

Part D

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

Express your answer to three significant figures and include the appropriate units.

Answer 1

Part C According to the law of conservation of momentum. Forx-component of the momentum, For y-component of the momentum, + m,v^ sin6, .. (2) From equation(1), mv sin....3) sin θ Using equation (3) in equation (1),

-2.10x sin(30) 2.80-2.10xcos (30) - tan 46.91° Hence the angle O, is 4691 Part D From equation (3) sin θ -0.125 x2.10x sin (30) 0.140xsin(-46.91) -1.283 Hence the speed vj is 1.283