A ball moving with a velocity v collides head | vedclass

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(B) Given: Mass of both balls is $m$. Initial velocities are $u_1 = v$ and $u_2 = 0$. After collision, $v_1 = 0.15 v$.Using the law of conservation of

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$25$

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(B) Given: Mass of both balls is $m$. Initial velocities are $u_1 = v$ and $u_2 = 0$. After collision, $v_1 = 0.15 v$.Using the law of conservation of linear momentum:$m u_1 + m u_2 = m v_1 + m v_2$$m v + 0 = m(0.15 v) + m v_2$$v = 0.15 v + v_2 \implies v_2 = 0.85 v$Initial kinetic energy $(KE)_i = \frac{1}{2} m v^2$.Final kinetic energy $(KE)_f = \frac{1}{2} m v_1^2 + \frac{1}{2} m v_2^2 = \frac{1}{2} m (0.15 v)^2 + \frac{1}{2} m (0.85 v)^2$$(KE)_f = \frac{1}{2} m v^2 [0.15^2 + 0.85^2] = \frac{1}{2} m v^2 [0.0225 + 0.7225] = \frac{1}{2} m v^2 [0.745]$Decrease in kinetic energy $\Delta KE = (KE)_i - (KE)_f = \frac{1}{2} m v^2 [1 - 0.745] = 0.255 	imes (KE)_i$Percentage decrease $= \frac{\Delta KE}{(KE)_i} 	imes 100 = 0.255 	imes 100 = 25.5 \% \approx 25 \%$.

$A$ ball moving with a velocity $v$ collides head-on with a stationary second ball of the same mass. After the collision, the velocity of the first ball is reduced to $0.15 v$. The kinetic energy of the system is decreased nearly by (in $\%$)

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