Chapter: Kinetics of particle· Engineering Mechanics

A cricket ball of mass 0.16 kg moving horizontally at 25 m/s is hit straight back by a bat and leaves at 35 m/s. If the bat is in contact with the ball for 0.02 s, find the impulse and the average force exerted by the bat.

Q.16 A cricket ball of mass 0.16 kg moving horizontally at 25 m/s is hit straight back by a bat and leaves at 35 m/s. If the bat is in contact with the ball for 0.02 s, find the impulse and the average force exerted by the bat.

Doubtly AI Asked 27 Oct 2026 AI-generated · not yet verified Practice test mu engineering Save WhatsApp Report

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Answer: Impulse \(=9.6\) N s; average force \(=480\) N (both directed opposite to the ball's initial motion).

Take the direction of the returning ball as positive. Initial velocity \(u=-25\) m/s, final velocity \(v=+35\) m/s.

Impulse–momentum principle: impulse \(=\) change in momentum

\[J=m(v-u)=0.16\,[35-(-25)]=0.16\times60=9.6\ \text{N s}\]

Average force: \(F_{avg}=\dfrac{J}{\Delta t}=\dfrac{9.6}{0.02}=480\) N.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

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A cricket ball of mass 0.16 kg moving horizontally at 25 m/s is hit straight back by a bat and leaves at 35 m/s. If the bat is in contact with the ball for 0.02 s, find the impulse and the average force exerted by the bat.

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