Question

Billiard ball A of mass mA = 0.130 kg moving with speed vA = 2.95 m/s strikes ball B, initially at rest, of mass mB = 0.150 kg . As a result of the collision, ball A is deflected off at an angle of 29.5 ∘ with a speed vA1 = 2.25 m/s

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

Express your answer numerically with the appropriate units.

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

Express your answer numerically with the appropriate units.

Answer 1

conservation of linear momentum along horizontal

0.13* 2.95 = 0.13* 2.25* cos 29.5 + 0.15* v cos x

v cos x = 0.8595 ... (i)

conservation of linear momentum along vertical

v sin x = 0.13* 2.25* sin 29.5

v sin x = 0.144 .. (ii).

square and adding (i) and (ii)

v = 0.8715 m/s

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b)

putting in (i)

x = 9.513 deg

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