Question

A ball of mass m moving at 5 m/s at 0° bounces off an identical ball of mass m at rest. The first ball bounces off the second ball and moves away at a velocity of 2.5 m/s and an angle of -60°. If the second ball is moving at 4.36 m/s after the cos m bounce, what is its angle? SrM 0

Answer 1

Using momentum conservation

In horizontal direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*V1x + m2*V2x

m1 = m2 = mass of each ball = m

u1x = initial velocity of moving ball in horizontal direction = 5 m/sec

u2x = initial velocity of stationary ball in horizontal direction = 0 m/sec

V1x = final velocity of moving ball in horizontal direction = 2.5*cos (-60 deg) = 1.25 m/sec

V2x = final velocity of moving ball in horizontal direction

Given that V2 = 4.36 m/sec

Suppose direction of final ball is $\theta$, then

V2x = V2*cos $\theta$ = 4.36*cos $\theta$

Using these values:

m*5 + m*0 = m*1.25 + m*4.36*cos $\theta$

5 + 0 = 1.25 + 4.36*cos $\theta$

cos $\theta$ = (5 - 1.25)/4.36

$\theta$ = arccos (3.75/4.36) = +30.7 deg

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