A ball travelling at $15 m / s$ is returned by a racket in the opposite direction with velocity of $20 m / s$ .If the change in KE of the ball thereby is $8.75 J$ ,find the change in its momentum ( $3.5 k g m / s$ )

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Solution

From the figure,

$v_{i} = - 15 m / s$

$v_{f} = 20 m / s$

$K E = \frac{1}{2} m v^{2}$

$\frac{1}{2} m ({v_{f}}^{2} - {v_{i}}^{2}) = 8.75$

$\frac{1}{2} m (v_{f} - v_{i}) (v_{f} + v_{i}) = 8.75$

$\frac{m}{2} (v_{f} - v_{i}) (20 + (- 15)) = 8.75$

$∵ c h a n g e i n m o m e n t u m = m (v_{f} - v_{i})$

$Δ p = 3.5 K g m / s$

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A ball travelling at 15m/s is returned by a racket in the opposite direction with velocity of 20m/s.If the change in KE of the ball thereby is 8.75J,find the change in its momentum (3.5kgm/s)

From the figure, vi=−15m/s vf=20m/s KE=12mv2 12m(vf2−vi2)=8.75 12m(vf−vi)(vf+vi)=8.75 m2(vf−vi)(20+(−15))=8.75 ∵changeinmomentum=m(vf−vi) Δp=3.5Kgm/s

A ball travelling at $15 m / s$ is returned by a racket in the opposite direction with velocity of $20 m / s$ .If the change in KE of the ball thereby is $8.75 J$ ,find the change in its momentum ( $3.5 k g m / s$ )