Question:
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
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
- \(0.07\;N\)
- \(0.05\;N\)
- \(5\;N\)
- \(7\;N\)
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The Correct Option is B
Solution and Explanation
Step 1: Convert the given quantities into SI units.
Mass of the ball: \[ m=50\;g=\frac{50}{1000}\;kg \] \[ m=0.05\;kg \] Initial velocity: \[ u=50\;\text{cm s}^{-1} \] \[ u=\frac{50}{100}\;\text{m s}^{-1} \] \[ u=0.5\;\text{m s}^{-1} \] Final velocity: \[ v=0 \] Time taken: \[ t=0.5\;s \]
Step 2: Find the acceleration.
Using the equation \[ a=\frac{v-u}{t} \] Substituting the values, \[ a=\frac{0-0.5}{0.5} \] \[ a=-1\;\text{m s}^{-2} \] The negative sign indicates retardation.
Step 3: Apply Newton's second law.
Force is given by \[ F=ma \] \[ F=0.05\times(-1) \] \[ F=-0.05\;N \] Magnitude of the force is \[ 0.05\;N \]
Step 4: Final conclusion.
Hence, the force applied to stop the ball is \[ \boxed{0.05\;N} \]
Mass of the ball: \[ m=50\;g=\frac{50}{1000}\;kg \] \[ m=0.05\;kg \] Initial velocity: \[ u=50\;\text{cm s}^{-1} \] \[ u=\frac{50}{100}\;\text{m s}^{-1} \] \[ u=0.5\;\text{m s}^{-1} \] Final velocity: \[ v=0 \] Time taken: \[ t=0.5\;s \]
Step 2: Find the acceleration.
Using the equation \[ a=\frac{v-u}{t} \] Substituting the values, \[ a=\frac{0-0.5}{0.5} \] \[ a=-1\;\text{m s}^{-2} \] The negative sign indicates retardation.
Step 3: Apply Newton's second law.
Force is given by \[ F=ma \] \[ F=0.05\times(-1) \] \[ F=-0.05\;N \] Magnitude of the force is \[ 0.05\;N \]
Step 4: Final conclusion.
Hence, the force applied to stop the ball is \[ \boxed{0.05\;N} \]
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