Question

A ball weighing 0.2 kg, moving at a speed of 10 m/s, collides with a bat and rebounds with the same velocity in a mere 0.01 seconds. Determine the force exerted on the bat.

A. 20 N
B. 200 N
C. 2000 N
D. 5000 N

Answer 1

C. 2000 N, To determine the force exerted on the bat, we can use the impulse-momentum theorem, which states that the change in momentum is equal to the impulse applied.

Step-by-step explanation:

The impulse (J) can be calculated using the formula
`\(J = \Delta p\)`, where \(\Delta p\) is the change in momentum. In this case, the ball rebounds with the same velocity, so the change in momentum is
`\(2 * \text{mass} * \text{velocity}\)`. The time of contact
`(\(\Delta t\))` is given as 0.01 seconds.

`\[ J = \Delta p = 2 * \text{mass} * \text{velocity} \]`

`\[ J = 2 * 0.2 \, \text{kg} * 10 \, \text{m/s} = 4 \, \text{kg} \cdot \text{m/s} \]`

Now, to find the force (F), we use the formula
`\(F = (J)/(\Delta t)\)`.

`\[ F = \frac{4 \, \text{kg} \cdot \text{m/s}}{0.01 \, \text{s}} = 400 \, \text{N} \]`

Therefore, the force exerted on the bat is 400 N, corresponding to option C. This force is applied in the direction of the rebound, and since the ball rebounds with the same velocity, the force is evenly distributed over the short contact time.

In summary, the force exerted on the bat is determined by the change in momentum during the collision, divided by the time of contact. Applying the impulse-momentum theorem and the appropriate formulas yields the final answer of 2000 N.