In a carrom game, a striker of mass 15 g hits a coin of mass 5 g head on and the coin moves with a speed of 0.36 m/s. If the time of contact between the striker and the coin is 3 milliseconds, then what is the average force applied by the striker on the coin?
\(0.60\ N\)
By the impulse-momentum theorem applied to the coin (initially at rest), \(F \times t = \Delta p = m_{coin} v\).
Here \(m_{coin} = 5\ g = 0.005\ kg\), \(v = 0.36\ m/s\) and \(t = 3\ ms = 0.003\ s\).
So \(F = \dfrac{m_{coin} v}{t} = \dfrac{0.005 \times 0.36}{0.003} = \dfrac{0.0018}{0.003} = 0.6\ N\). Hence option (b) is correct.
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