Billiard ball A of mass m A = 0.123 kg moving with speed v A = 2.80 m/s strikes

Question

Billiard ball A of mass mA = 0.123 kg moving with speedvA = 2.80 m/s strikes ball B, initially at rest, of mass mB= 0.138 kg . As a result of the collision, ball A is deflected off at an angle of A = 30.0 with a speed vA = 2.10 m/s,and ball B moves with a speed vB at an angle of B to original direction of motion of ball A.

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

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

(0=mAvAsinAmBvBsinB)

(mAvA=mAvAcosA+mBvBcosB)

Explanation / Answer

(a)

. given data

m(a)=.123kg

v(a)=2.80m/s

m(b.138kg

thita=30 degree

v(a)=2.10m/s

P = 0

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

0.123 (2.8){1,0} - 0.123 (2.1)(cos 30, sin 30) - vB' (0.13) (cos , sin ) = 0



(b).

0.123 (2.8){1,0} - 0.12 (2.1)(cos 30, sin 30) - vB' (0.13) (cos , sin ) = 0

0.12 (2.8) - 0.12 (2.1) cos 30 = vB' (0.14) cos . . . . . . .(eqs 1)

0 - 0.12 (2.1) sin 30 = vB' (0.13) sin . . . . . . .(eqs 2)

equation 2 divided by equation 1,

tan = (0 - 0.12 (2.1) sin 30°)/(0.12 (2.8) - 0.12 (2.1) cos 30°)

= -46.94°

back to equation 2,

0 - 0.12 (2.1) sin 30° = vB' (0.13) sin (-46.94°)

vB' = 1.2218 m/s

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