Question

A carrom coin of radius 1.5 cm and mass 5 g strikes the side of the board at anangle 30° with respect to the normal. The coin stays in contact with the side for(10^-3)s and rebounds at an angle of 45° with respect to the normal. The initialspeed of the coin is 1 m/s and its final speed is 0.7m/s. The average impulsiveforce on the coin in the normal direction is:​

Answer 1

initial velocity of carrom coin along normal ,u = 1m/s × cos30°

= √3/2 = 0.866 m/s

final velocity of carrom coin along normal, v =- 0.7 m/s × cos45° = -0.7/1.4 = -0.5m/s [ here negative sign indicates that Initial velocity of coin along normal is just opposite to final velocity of coin along normal]

time taken to stay in contact with the side of board, t = 10^-3 sec

so, change in momentum during this time = m(v - u)

= 5g × {0.866 -(-0.5)}m/s

= 5 × 10^-3 × 1.366 Kgm/s

now impulsive force = change in momentum/time

= 5 × 10^-3 × 1.366/10^-3 N

= 5 × 1.366 N

= 6.83 N