Question

A cricket ball of mass \(150\,\text{g}\) moving with a velocity of \(12\,\text{m/s}\) is turned back with a velocity of \(20\,\text{m/s}\) on hitting the bat. The force of the blow lasts for \(0.01\,\text{s}\). The force exerted on the ball by the bat is

• A. \(480\,\text{N}\)
• B. \(240\,\text{N}\)
• C. \(360\,\text{N}\)
• D. \(120\,\text{N}\)

Hint:

Always take care of direction while calculating change in velocity.

Answer 1

The Correct Option is A

Step 1: Converting mass into kg.
\[ m = 150\,\text{g} = 0.15\,\text{kg} \]
Step 2: Calculating change in velocity.
Initial velocity \(u = 12\,\text{m/s}\)
Final velocity \(v = -20\,\text{m/s}\)
\[ \Delta v = v - u = -20 - 12 = -32\,\text{m/s} \]
Step 3: Change in momentum.
\[ \Delta p = m \Delta v = 0.15 \times 32 = 4.8\,\text{kg m/s} \]
Step 4: Force exerted.
\[ F = \frac{\Delta p}{\Delta t} = \frac{4.8}{0.01} = 480\,\text{N} \]
Step 5: Conclusion.
The force exerted on the ball is \(480\,\text{N}\).