Question

Billiard ball A of mass mA = 0.130 kg moving with speed vA = 2.65 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 30.5 ∘ with a speed vA1 = 2.35 m/s .

C.) Solve for the speed, vB1, of ball B. Do not assume the collision is elastic.

D.) Solve for the angle, θB, of ball B. Do not assume the collision is elastic.

Answer 1

from conservation of momentum in X direction:

mA vA = mA vA' cos(thetaA)A + mB VB' cos(thetaB)

along y direction:

0 = mA V'A sin(thetaA) + mB vB' sin(thetaB)solving the above eqn for vB' we get

vB' = -(mA VA' sin(thetaA))/mB sin(thetaB)

putting the value of vB' in X eqn

mA vA = mA vA' cos(thetaA) + mB -(mA VA' sin(thetaA))/mB sin(thetaB) x cos(thetB)

tan(thetaB) = -vA' sin(thetaA)/(vA - vA' cos(thetaA))

tan(thetaB) = -2.35 x sin30.5/(2.65 - 2.35 x cos30.5) => thetaB = -62.34 deg

Hence, thetaB = -62.34 deg <<<<<<<<<<<< Part(D)

vB' = -(mA VA' sin(thetaA))/mB sin(thetaB)

vB' = -(0.13 x 2.35 sin30.5)/(0.14 x sin-62.34) = 1.25 m/s

Hence, vB' = 1.25 m/s <<<<<<<<< Part(C)