Question:easy

A cricket ball of mass \(50\;g\) having velocity \(50\;\text{cm s}^{-1}\) is stopped in \(0.5\;s\). The force applied to stop the ball is

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Always convert quantities into SI units before applying formulas in mechanics problems.
Updated On: Jun 22, 2026
  • \(0.07\;N\)
  • \(0.05\;N\)
  • \(5\;N\)
  • \(7\;N\)
Show Solution

The Correct Option is B

Solution and Explanation

Step 1: Collect the data and convert to SI units.
Mass $m = 50\ g = 0.05\ kg$, initial speed $u = 50\ \text{cm s}^{-1} = 0.5\ \text{m s}^{-1}$, final speed $v = 0$ (it stops), and time $t = 0.5\ s$. We want the stopping force.
Step 2: Decide on the approach.
Force relates to the rate of change of momentum, so the impulse-momentum idea is the cleanest path here.
Step 3: Find the change in momentum.
\[ \Delta p = m(v - u) = 0.05(0 - 0.5) = -0.025\ \text{kg m s}^{-1} \]
Step 4: Relate force to momentum change.
By Newton's second law in impulse form, \[ F = \frac{\Delta p}{t} = \frac{-0.025}{0.5} \]
Step 5: Compute the value.
\[ F = -0.05\ N \]
Step 6: Interpret the result.
The negative sign just shows the force opposes the motion; its magnitude is $0.05\ N$, matching option (2). \[ \boxed{0.05\ N} \]
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