A 100, g iron ball having velocity 10, m/s co | vedclass

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(A) Given: Mass $m = 100\, g = 0.1\, kg$,velocity $v = 10\, m/s$,time of contact $\Delta t = 0.1\, s$,angle with the wall $	heta = 30^\circ$.The comp

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$10$

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(A) Given: Mass $m = 100\, g = 0.1\, kg$,velocity $v = 10\, m/s$,time of contact $\Delta t = 0.1\, s$,angle with the wall $	heta = 30^\circ$.The component of momentum perpendicular to the wall changes during the collision. The angle of the velocity vector with the normal to the wall is $\phi = 90^\circ - 30^\circ = 60^\circ$.The change in momentum $\Delta p$ is given by the change in the component of momentum perpendicular to the wall:$\Delta p = p_f - p_i = (-mv \cos \phi) - (mv \cos \phi) = -2mv \cos \phi$.The magnitude of the change in momentum is $|\Delta p| = 2mv \cos 60^\circ$.Force $F = \frac{|\Delta p|}{\Delta t} = \frac{2mv \cos 60^\circ}{\Delta t}$.Substituting the values:$F = \frac{2 	imes 0.1\, kg 	imes 10\, m/s 	imes \cos 60^\circ}{0.1\, s} = \frac{2 	imes 0.1 	imes 10 	imes 0.5}{0.1} = 10\, N$.

$A$ $100\, g$ iron ball having velocity $10\, m/s$ collides with a wall at an angle of $30^\circ$ with the wall 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 wall is ........... $N$.

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