Question

A cricket ball of mass $150{\text{g}}$ is moving with a velocity of $12{\text{m}}{{\text{s}}^{ - 1}}$ . It is hit by a bat and turns back with a velocity of $20{\text{m}}{{\text{s}}^{ - 1}}$ . If the force of the blow acts for $0.001{\text{s}}$ on the ball, find the force.
A) 4800 N
B) 5200 N
C) 6000 N
D) 7000 N

Answer 1

Hint: The force of the blow acts for a very short period and causes a change in the momentum of the ball as its velocity changes. This change in momentum of the ball is referred to as the impulse of the ball. The impulse of the ball can be expressed as the product of the force acting and the time for which it acts. The force can thus be obtained by calculating the impulse.

Formula Used:
1) Impulse acting on an object is given by, $J = \Delta p$ or $J = Ft$ where $\Delta p$ is the change in momentum of the object, $F$ is the force acting and $t$ is the time for which it acts.
2) The change in momentum of an object is expressed as $\Delta p = mv - mu$ where $m$ is the mass of the object, $v$ is its final velocity and $u$ is its initial velocity.

Complete step by step answer:
Step 1: List the parameters known from the question.
The mass of the ball is $m = 150{\text{g}} = 0.15{\text{kg}}$ and its initial velocity is $u = 12{\text{m}}{{\text{s}}^{ - 1}}$ . When the ball gets hit by the bat its velocity changes to $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ .
The impulsive force acts on the ball for a time $t = 0.001{\text{s}}$
Step 2: Find the impulse acting on the ball.
The impulse-momentum relation is given by, $J = \Delta p$ -------- (1) where $\Delta p$ is the change in momentum of the object.
The change in momentum of the ball will be the difference between its final momentum and its initial momentum.
i.e., $\Delta p = mv - mu$ -------- (2) where $m$ is the mass of the ball, $v$ is its final velocity and $u$ is its initial velocity.
Substituting values for $m = 0.15{\text{kg}}$ , $u = - 12{\text{m}}{{\text{s}}^{ - 1}}$ and $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ in equation (2) we get, $\Delta p = \left( {0.15 \times 20} \right) - \left( {0.15 \times \left( { - 12} \right)} \right) = 4.8{\text{kgm}}{{\text{s}}^{ - 1}}$
We obtain the change in momentum of the ball as $\Delta p = 4.8{\text{kgm}}{{\text{s}}^{ - 1}}$ .
Then from equation (1), we find that the impulse of the ball is $J = \Delta p = 4.8{\text{kgm}}{{\text{s}}^{ - 1}}$
Step 3: Find the force acting on the ball using the concept of impulse.
When the ball gets hit by the bat we say the bat has imparted impulse to the ball. Impulse can be defined as the integral of the force over the time interval for which it acts. Hence impulse can also be expressed as $J = Ft$
Then from the above relation, we can express the force as $F = \dfrac{J}{t}$ ------ (3)
Substituting values for $J = 4.8{\text{kgm}}{{\text{s}}^{ - 1}}$ and $t = 0.001{\text{s}}$ in equation (3) we get, $F = \dfrac{{4.8}}{{0.001}} = 4800{\text{N}}$
So, the force of the blow is $F = 4800{\text{N}}$ and thus the correct option is A.

Note: As the ball encounters the bat, its velocity changes to $v = 20{\text{m}}{{\text{s}}^{ - 1}}$ and the direction of motion must have also changed so that it moves in the opposite direction. Here we considered the initial direction of the ball to be along the negative x-direction and so $u = - 12{\text{m}}{{\text{s}}^{ - 1}}$ is substituted in equation (2). If we considered it as the positive x-direction, the force would simply have a negative sign.