A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a rate of $20 \mathrm{~m} / \mathrm{s}$.…
A player caught a cricket ball of mass $150 \mathrm{~g}$ moving at a rate of $20 \mathrm{~m} / \mathrm{s}$. If the catching process is completed in $0.1 \mathrm{~s}$, the force of the blow exerted by the ball on the hand of the player is equal to
$300 \mathrm{~N}$
$150 \mathrm{~N}$
$3 \mathrm{~N}$
$30 \mathrm{~N}$
Solution
$(m v-0) \Rightarrow 0.15 \times 20$
$\mathrm{F}=\frac{3}{0.1}=30 \mathrm{~N}$
$
The force exerted by the ball on the hand of the player can be calculated using the formula for force, which is change in momentum divided by the time taken.
The momentum of the ball before being caught is given by the mass of the ball multiplied by its velocity, which is 150g * 20m/s = 3000g*m/s. Because the player catches the ball, its final velocity becomes 0, so the final momentum is 0.
The change in momentum is therefore 3000g*m/s - 0 = 3000g*m/s.
However, we need to convert this into kg*m/s. So it is 3 kg*m/s
Now, Force = Change in momentum / Time taken
Force = 3 kg*m/s / 0.1 s = 30 N
$
Therefore, the force of the blow exerted by the ball on the hand of the player is 30 Newtons