A 0,34-kg volleyball is flying west at 1,98 m/s when it strikes a stationary 0,60-kg basketball dead centre. The volleyball rebounds east at 0,79 m/s. What will be the velocity of the basketball immediately after impact? Choose the east direction as positive.

Explanation:

use the law of conservation of momentum

M1U1+M2U2=M1V1+M2V2

m-mass

u-initial velocity

v-final velocity

1-volleyball

2-basket ball dead center

so using the above formula

0.34×1.98+0.64×0=0.34×0.79+0.64×V2

to get

0.6732+0 = 0.2686+0.64V2

0.4046=0.64V2

divide both sides by 0.64 to get V2 which is the final velocity of basketball

0.6322m/s=V2

i think the answer would be b

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