A 100 , g iron ball having a velocity of 10 , | vedclass

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(B) The force experienced by the ball is equal to the rate of change of momentum.Given: mass $m = 100 \, g = 0.1 \, kg$,velocity $v = 10 \, m/s$,angle

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$10$

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(B) The force experienced by the ball is equal to the rate of change of momentum.Given: mass $m = 100 \, g = 0.1 \, kg$,velocity $v = 10 \, m/s$,angle with the wall $	heta = 30^\circ$,and time $t = 0.1 \, s$.The change in momentum $\Delta p$ is given by the component perpendicular to the wall: $\Delta p = 2mv \sin 	heta$.Substituting the values: $\Delta p = 2 	imes 0.1 	imes 10 	imes \sin 30^\circ = 2 	imes 0.1 	imes 10 	imes 0.5 = 1 \, kg \cdot m/s$.The force $F = \frac{\Delta p}{t} = \frac{1}{0.1} = 10 \, N$.

$A$ $100 \, g$ iron ball having a velocity of $10 \, m/s$ collides with a wall at an angle of $30^\circ$ with the wall surface and rebounds with the same angle. If the period of contact between the ball and the wall is $0.1 \, s$,then the force experienced by the ball is ............. $N$.

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