Question

A player caught a cricket ball of mass 150g moving with $20m{s^{ - 1}}$. If the catching process is completed in 0.1 sec.
A) Force exerted by the ball on the hand of the player is 30N.
B) Force exerted by the ball on the hand of the player is 40N.
C) Impulse of the ball is 3N-S.
D) Force exerted by the ball on the hand of player id 50N.

Answer 1

Hint: Force is defined as the product of mass and acceleration of the body; it is also defined as the rate of change of momentum with respect to time. The impulse is the product of the force applied on the body and time interval for which the force is applied on the body.
Formula used: The formula of the force is given by,
$F = m \cdot a$
Where force is $F$ mass of the body is $m$ and acceleration of the body is $a$.
The formula of the impulse is given by,
$I = F \cdot t$
Where impulse is $I$ force on the body is $F$ and the time taken is $t$.
The formula of the momentum is given by,
$p = mv$
Where momentum is $p$ mass of the body is $m$ and velocity is $v$.

Complete step by step answer:
It is given in the problem that a player caught a cricket ball of mass 150g moving with$20m{s^{ - 1}}$ the catching process is completed in 0.1 sec. and we need to find the force applied on the hands of the player and the impulse on the hand.
The formula of the momentum is given by,
$p = mv$
Where momentum is $p$ mass of body is $m$ and velocity is $v$.
Therefore the momentum is,
$ \Rightarrow p = mv$
Since the mass of the body is 0.15kg with initial speed $20m{s^{ - 1}}$ and final speed is zero.
Replace the value of the mass and initial velocity.
$ \Rightarrow p = \left( {\dfrac{{150}}{{1000}}} \right) \times \left( {20} \right)$
$ \Rightarrow p = 3kg - m{s^{ - 1}}$
Let us calculate value of acceleration,
$ \Rightarrow a = \dfrac{{v - u}}{t}$
$ \Rightarrow a = \dfrac{{0 - 20}}{{0.1}}$
$ \Rightarrow a = - 200m{s^{ - 2}}$
The formula of the force is given by,
$F = m \cdot a$
Where force is $F$ mass of the body is $m$ and acceleration of the body is $a$.
Since the mass of the body is 0.15kg with acceleration $a = - 200m{s^{ - 2}}$ therefore we get,
$ \Rightarrow F = m \cdot a$
$ \Rightarrow F = \left( {0.15} \right) \times \left( { - 200} \right)$
$ \Rightarrow F = - 30N$
So the magnitude of force is given by $F = 30N$.
Now let us calculate the impulse.
The formula of the impulse is given by,
$I = F \cdot t$
Where impulse is $I$ force on the body is $F$ and the time taken is $t$.
The force on the body is $F = 30N$ and the time taken is 0.1sec. therefore we get,
$ \Rightarrow I = F \cdot t$
$ \Rightarrow I = \left( {30} \right) \times \left( {0.1} \right)$
$ \Rightarrow I = 3N - s$

The impulse is $I = N - s$. Therefore the force on the hand of the player is $F = 30N$ and impulse is $I = 3N - s$. The correct answer for the problem is option A and option C.

Note:
Momentum is defined as the product of mass and velocity: its unit is $kg - m{s^{ - 1}}$. The negative sign on the force represents that the applied force is opposite to the direction of motion as the players hand are resisting the motion of the ball.