A cricket ball of mass 150 gram moving with a velocity of 12 m/s strikes against the bat. It rebounces with a velocity of 20 m/s. The ball remains in touch with the batfor 0.01 second. The average force applied by the bat on the ball is

840 Newton

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480 Newton

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804 Newton

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408 Newton

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Solution

The correct option is B 480 Newton
When the direction of motion is reversed,
change in momentum $= m (v + u)$
$∴ F o r c e = \frac{m (υ + u)}{t}$
$= \frac{15 \times 10^{- 2} \times 32}{10^{- 2}} = 480$

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A cricket ball of mass 150 gram moving with a velocity of 12 m/s strikes against the bat. It rebounces with a velocity of 20 m/s. The ball remains in touch with the batfor 0.01 second. The average force applied by the bat on the ball is

The correct option is B 480 NewtonWhen the direction of motion is reversed,change in momentum =m(v+u)∴Force=m(υ+u)t=15×10−2×3210−2=480

The correct option is B 480 Newton
When the direction of motion is reversed,
change in momentum $= m (v + u)$
$∴ F o r c e = \frac{m (υ + u)}{t}$
$= \frac{15 \times 10^{- 2} \times 32}{10^{- 2}} = 480$