A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of $12 m s^{- 1}$ . If the mass of the ball is $0.15 k g$ , determine the change in momentum of the ball. (Assume linear motion of the ball)

3.6 k g m / s

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5 k g m / s

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3.5 k g m / s

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7 k g m / s

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Solution

The correct option is A $3.6 k g m / s$
The initial momentum of ball $P_{1} = m \to v$ , after batsman hits ball and reverses its direction, momentum of ball $P_{2} = - m \to v$
Magnitude of change in momentum $Δ p = | P_{2} - P_{1} | = | - 2 m \to v | = 2 m | \to v |$
from data given in question $m = 0.15 k g; | \to v | = 12 m / s$
$Δ P = 2 \times 0.15 \times 12 = 3. 6 k g m / s$

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