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A $100\, kg$ gun fires a ball of $1\, kg$ horizontally from a cliff of height $500 \,m $. It falls on the ground at a distance of $400 \,m $ from bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity $g = 10\,ms^{-1}$ )
Solution
Mass of gun $m_{1}=100 \mathrm{~kg}$
mass of ball $m_{2}=1 \mathrm{~kg}$
Horizontal distance travelled by the ball, $x=400 \mathrm{~m}$
Initially vertical component of velocity is zero (in downward direction)
$h=\frac{1}{2} g t^{2}$
$500=\frac{1}{2}(10) t^{2}$
$500=5 t^{2}$
$t^{2}=100$
$t=\sqrt{100}=10 \mathrm{~s}$
For horizontal motion of ball,
$x=u t$,
$u=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$
For hor $x=u t$ $t=\frac{x}{t}=\frac{400}{10}=40 \mathrm{~m} / \mathrm{s}$ Let $v$ is recoil velocity of gun,
By law of conservation of linear momentum, $m_{1} v=m_{2} u$ $\therefore v=\frac{m_{2} u}{m_{1}}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s}$
