A ball of mass 100 g moving with velocity of 10 $m s^{- 1}$ is stopped by a boy in 0.2 s. Calculate the force applied by the boy to stop the ball and the acceleration of the ball.

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Solution

Given : $m = 100 g = 0.1 k g$ $v = 10 m / s$ $Δ t = 0.2 s$
Change in momentum $Δ P = | P_{i} - P_{f} | = m v - 0$
$⟹ Δ P = 0.1 \times 10 = 1 k g m / s$
Force $F = \frac{Δ P}{Δ t} = \frac{1}{0.2} = 5 N$
Acceleration $a = \frac{F}{m} = \frac{5}{0.1} = 50 m / s^{2}$

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A ball of mass 100 g moving with velocity of 10 ms−1 is stopped by a boy in 0.2 s. Calculate the force applied by the boy to stop the ball and the acceleration of the ball.

Given : m=100 g=0.1 kg v=10 m/s Δt=0.2 sChange in momentum ΔP=|Pi−Pf|=mv−0⟹ ΔP=0.1×10=1 kg m/sForce F=ΔPΔt=10.2=5 NAcceleration a=Fm=50.1=50 m/s2

A ball of mass 100 g moving with velocity of 10 $m s^{- 1}$ is stopped by a boy in 0.2 s. Calculate the force applied by the boy to stop the ball and the acceleration of the ball.

Given : $m = 100 g = 0.1 k g$ $v = 10 m / s$ $Δ t = 0.2 s$
Change in momentum $Δ P = | P_{i} - P_{f} | = m v - 0$
$⟹ Δ P = 0.1 \times 10 = 1 k g m / s$
Force $F = \frac{Δ P}{Δ t} = \frac{1}{0.2} = 5 N$
Acceleration $a = \frac{F}{m} = \frac{5}{0.1} = 50 m / s^{2}$