Question

A tennis ball of mass 150 g is moving at $20~ms^{-1}$. A racket strikes it, reversing its direction with a final speed of $30~ms^{-1}$. If the contact time is 0.02 s, then the magnitude of the force (in N) exerted by the racket is ________.

• A. 1.5 N
• B. 3.75 N
• C. 15 N
• D. 150 N
• E. 375 N

Hint:

When a ball bounces back, change in velocity is the sum of speeds: $v_1 + v_2$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept
This problem involves the impulse-momentum theorem. The impulse delivered to an object (which is the average force multiplied by the contact time) is equal to the change in the object's momentum. We need to calculate the change in momentum and then use the contact time to find the average force.
Step 2: Key Formula or Approach
1. Momentum (\(p\)): \(p = mv\)
2. Change in Momentum (\(\Delta p\)): \(\Delta p = p_{final} - p_{initial} = mv_{final} - mv_{initial}\)
3. Impulse-Momentum Theorem: Impulse \(J = F_{avg} \cdot \Delta t = \Delta p\)
From this, the average force is \(F_{avg} = \frac{\Delta p}{\Delta t}\).
It's crucial to handle the vector nature of velocity correctly. Since the direction is reversed, one of the velocities must be taken as negative.
Step 3: Detailed Explanation
1. List the given values in SI units.
- Mass, \(m = 150 \text{ g} = 0.150 \text{ kg}\)
- Contact time, \(\Delta t = 0.02 \text{ s}\)
- Initial velocity, \(v_{initial}\). Let's define the initial direction as positive. So, \(v_{initial} = +20 \text{ ms}^{-1}\).
- Final velocity, \(v_{final}\). The direction is reversed, so it's in the negative direction. \(v_{final} = -30 \text{ ms}^{-1}\).
2. Calculate the change in momentum (\(\Delta p\)).
\[ \Delta p = m(v_{final} - v_{initial}) \] \[ \Delta p = 0.150 \text{ kg} \times (-30 \text{ ms}^{-1} - 20 \text{ ms}^{-1}) \] \[ \Delta p = 0.150 \times (-50) \] \[ \Delta p = -7.5 \text{ kg ms}^{-1} \] The negative sign indicates the direction of the change in momentum (and thus the force), which is opposite to the initial direction of motion.
3. Calculate the magnitude of the average force.
\[ F_{avg} = \frac{\Delta p}{\Delta t} \] \[ F_{avg} = \frac{-7.5 \text{ kg ms}^{-1}}{0.02 \text{ s}} \] \[ F_{avg} = -375 \text{ N} \] The question asks for the magnitude of the force, which is the absolute value.
\[ |F_{avg}| = 375 \text{ N} \] Step 4: Final Answer
The magnitude of the force exerted by the racket is 375 N.