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3 years ago

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In a pool game, the cue ball, which has an initial speed of 3.0 m/s, make an elastic collision with the eight ball, which is ini

tially at rest. After the collision, the eight ball moves at an angle of 40° to the original direction of the cue ball. (a) Find the direction of motion of the cue ball after the collision. ° (from the original line of motion) (b) Find the speed of each ball. Assume that the balls have equal mass. m/s (cue ball) m/s (eight ball)

1 answer:

zhenek [66]3 years ago

5 0

Explanation:

Given

initial speed(u)=3 m/s

mass of each ball is m

Let the cue ball is moving in x direction initially

In elastic collision Energy and momentum is conserved

Let u be the initial velocity and $v_1 , v_2$ be the final velocity of 8 ball and cue ball respectively

$\frac{mu^2}{2}+0=\frac{mv^2_1}{2}+\frac{mv^2_2}{2}$

The angle after which cue ball is deflected is given by

$\theta _1=90-40=50^{\circ}$

Conserving momentum in x direction

$mu=mv_1cos40+mv_2cos50$

$3=v_1cos40+v_2cos50$

Along Y axis

$0+0=v_1sin40-v_2sin50$

$v_1sin40=v_2sin50$

substitute the value of $v_1$

we get $v_2=1.912 m/s$

$v_1=2.27 m/s$

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