Billiard ball A of mass m A = 0.125 kg moving with speed v A = 3.00 m/s strikes

Question

Billiard ball A of mass mA = 0.125 kg moving with speed vA = 3.00 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 30.0 with a speed vA1 = 2.20 m/s

Part A

Taking the x axis to be the original direction of motion of ball A, write down the equation expressing the conservation of momentum for the components in the x direction. Assume the velocity of ball B after collision is vB1.

Express your answer symbolically in terms of the variables mA, mB, vA, vA1, vB1, A, and B.

ans:

mAvAmAvA1cos(A)mBvB1cos(B)

Part B

Write down the equation expressing the conservation of momentum for the components in the y direction.

ans:

mAvA1sin(A)mBvB1sin(B)

Part C

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.

vB1 = [? value] [?unit]

Part D

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

Express your answer numerically with the appropriate units.

mAvAmAvA1cos(A)mBvB1cos(B)

Explanation / Answer

Billiard ball A of mass mA = 0.125 kg

speed vA = 3.00 m/s

mass mB = 0.150 kg

ANGLE OF DEFLECTION= 30.0o

SPEED OF DFLECTION=>vA1 = 2.20 m/s

0.125 (3.00) - 0.125 (2.2) cos 30 = vB' (0.150) cos . . . . . . .(eqs 1)

0 - 0.125 (2.2) sin 30 = vB' (0.150) sin . . . . . . .(eqs 2)

equation 2 divided by equation 1,

tan = 1.00
= -45.13°

back to equation 2,

0 - 0.125 (2.2) sin 30 = vB' (0.150) sin (-45.13)

vB' = 1.2934 m/s

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