A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a hei

Question

A ball of mass 0.5 kg is dropped from a height of 40 m . The ball hits the ground and rises to a height of 10 m . The impulse imparted to the ball during its collision with the ground is (Take $g=9.8 \mathrm{~m} / \mathrm{s}^2$ )

Professor PC Thomas Classes · Answer

(c): Mass of ball, $m=0.5 \mathrm{~kg}$ \ $ h_1=40 \mathrm{~m} 	ext { and } h_2=10 \mathrm{~m} $ So, velocity of ball before hitting the ground \ $ v_1=\sqrt{2 g h_1}=\sqrt{2 	imes 9.8 	imes 40} \Rightarrow v_1=28 \mathrm{~m} / \mathrm{s} $ \ After hitting the ground, let final velocity of the ball be  \ $v_2$. $ 	herefore v_2=\sqrt{2 g h_2}=\sqrt{2 	imes 9.8 	imes 10} ; v_2=14 \mathrm{~m} / \mathrm{s} $ \ Change in momentum, $\Delta p=m\left[v_2-\left(-v_1ight)ight]$ $ =\frac{1}{2}[14-(-28)] $ $\Delta p=21 \mathrm{~kg} \Rightarrow$ Impulse, $\Delta p=21 \mathrm{Ns}$