Billiard ball A of mass m A = 0.122 k g moving with speed v A = 2.80 m / s strik

Question

Billiard ball A of mass mA = 0.122kg moving with speed vA = 2.80 m/s strikes ball B, initially at rest, of mass mB = 0.140kg . As a result of the collision, ball A is deflected off at an angle of??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 ??B to original direction of motion of ball A.Solve these equations for the angle,

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

v?B, of ball B after the collision. Do not assume the collision is elastic.

Explanation / Answer

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

0.122 * (2.8) {1,0} - 0.122 * (2.10)*(cos 30, sin 30) - vB' (0.140) (cos (theta), sin (theta)) = 0

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

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

equation 2 divided by equation 1,

tan (theta) = (0 - 0.122 (2.10) sin 30

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