
A cricket ball of mass $250g$ collides with a bat with velocity $10\text{ m/s}$ and returns with the same velocity in 0.01 seconds. The force acted on the bat is:
A. $25\text{ N}$
B. $50\text{ N}$
C. $250\text{ N}$
D. $500\text{ N}$
Answer
596.1k+ views
Hint: Force acting on a body is defined as the rate of change of momentum acting on the body. So since the ball is moving initially with a velocity, it has an initial momentum which is applied on the bat when it strikes the bat. There is a change in the momentum when the ball leaves the bat after hitting it. So if we know the time period during this change we can calculate the force.
Complete Step-by-Step solution:
So the mass of the ball in the problem is $250g$ and the initial velocity of the ball before it hits the bat is $10\text{ m/s}$. After hitting the bat the ball is said to travel with a velocity of $10\text{ m/s}$ .
So the initial momentum is given by, ${{P}_{i}}=m{{v}_{i}}$, where ${{v}_{i}}$ is the initial velocity.
${{P}_{i}}=\left( 0.25kg \right)\times \left( 10m/s \right)$
${{P}_{i}}=2.5\text{ }kgm/s$ …equation (1)
The final momentum of the ball is given by, ${{P}_{f}}=m{{v}_{f}}$, where ${{v}_{f}}$ is the final velocity.
${{P}_{f}}=\left( 0.25kg \right)\times \left( 10m/s \right)$
${{P}_{f}}=2.5\text{ }kgm/s$ …. equation(2)
So, from equations (1) and (2), we can see that the magnitude of initial and the final momentum is equal but the direction is opposite to each other. So, we can write, ${{P}_{f}}=-{{P}_{i}}$.
So the rate of change of momentum in the time period of 0.01 seconds is the force acting on the bat, so we can write,
$\text{Force }(F)=\dfrac{{{P}_{f}}-{{P}_{i}}}{t}$
$F=\dfrac{2{{P}_{f}}}{t}\text{ }\left( \because {{P}_{i}}=-{{P}_{f}} \right)$
$\therefore F=\dfrac{2\times \left( 2.5kgm/s \right)}{0.01s}$
$F=500\text{ N}$
So the force acting on the bat is, $F=500\text{ N}$.
So the answer to the question is option(D).
Note: Momentum is a vector quantity which has both magnitude and direction. It can be defined as mass in motion. In a closed system (i.e. a system of particles on which no external force acts), the momentum is said to be conserved.
Momentum is conserved in all types of collisions, whether it be elastic or inelastic type of collision.
Complete Step-by-Step solution:
So the mass of the ball in the problem is $250g$ and the initial velocity of the ball before it hits the bat is $10\text{ m/s}$. After hitting the bat the ball is said to travel with a velocity of $10\text{ m/s}$ .
So the initial momentum is given by, ${{P}_{i}}=m{{v}_{i}}$, where ${{v}_{i}}$ is the initial velocity.
${{P}_{i}}=\left( 0.25kg \right)\times \left( 10m/s \right)$
${{P}_{i}}=2.5\text{ }kgm/s$ …equation (1)
The final momentum of the ball is given by, ${{P}_{f}}=m{{v}_{f}}$, where ${{v}_{f}}$ is the final velocity.
${{P}_{f}}=\left( 0.25kg \right)\times \left( 10m/s \right)$
${{P}_{f}}=2.5\text{ }kgm/s$ …. equation(2)
So, from equations (1) and (2), we can see that the magnitude of initial and the final momentum is equal but the direction is opposite to each other. So, we can write, ${{P}_{f}}=-{{P}_{i}}$.
So the rate of change of momentum in the time period of 0.01 seconds is the force acting on the bat, so we can write,
$\text{Force }(F)=\dfrac{{{P}_{f}}-{{P}_{i}}}{t}$
$F=\dfrac{2{{P}_{f}}}{t}\text{ }\left( \because {{P}_{i}}=-{{P}_{f}} \right)$
$\therefore F=\dfrac{2\times \left( 2.5kgm/s \right)}{0.01s}$
$F=500\text{ N}$
So the force acting on the bat is, $F=500\text{ N}$.
So the answer to the question is option(D).
Note: Momentum is a vector quantity which has both magnitude and direction. It can be defined as mass in motion. In a closed system (i.e. a system of particles on which no external force acts), the momentum is said to be conserved.
Momentum is conserved in all types of collisions, whether it be elastic or inelastic type of collision.
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