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Question
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 the bottom of the cliff. Find the recoil velocity of the gun. (acceleration due to gravity = 10 ms–2)
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Solution
Given, the mass of the gun (m1) = 100
Mass of the ball (m2) = 1 kg
Height of the cliff = 500 m
Horizontal distance travelled by the ball (x) = 400 m
From `h = 1/2 "gt"^2` .....(∵ Initial velocity in the downward direction is zero)
`500 = 1/2 xx 10 t^2`
`t = sqrt(100)` = 10 s
From `x = ut, u = x/t = 400/10` = 40 m/s
It v is the recoil velocity of gun, then according to the principle of conservation of linear momentum,
m1v = m2u
v = `(m_2u)/m_1 = 1/100 xx 40` = 0.4 m/s
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