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Q. A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $12 ms^{-1}$. If the mass of the ball is 0.15 kg, the imparted to the ball is

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Solution:

The initial momentum of ball $P _1= m \overrightarrow{ v }$, after batsman hits ball and reverses its direction, momentum of ball
$P _2=- m \overrightarrow{ v }$
Magnitude of change in momentum $\Delta p =\left| P _2- P _1\right|=|-2 m \overrightarrow{ v }|=2 m |\overrightarrow{ v }|$ from data given in question $m =0.15 kg ;|\overrightarrow{ v }|=12 m / s$
$ \Delta P =2 \times 0.15 \times 12=3.6 kgm / s $