A 100kg gun fires a ball of 1kg horizontally from a cliff of height 500m.It falls on the ground at a distance of 400m from the bottom of the cliff.Find the recoil velocity of the gun
Hi student,
Mass of the gun,
M = 100kg
Mass of the ball,
m = 1kg
Height of the cliff,
h = 500m
g = 10ms^−2
Time taken by the ball to reach the ground is
t =
√
2h/g = √2 × 500m/10ms^−2 = 10s
Horizontal distance covered = ut
i.e. 400×u×10
where
u
is the velocity of the ball.
u = 10ms^−1
According to law of conservation of linear momentum, we get
0 = Mv + μ
v = −mv/M = −(1kg)(40ms−1)/100kg = −0.4ms^−1
-ve sign shows that the direction of recoil of gun is opposite.
Feel free to ask doubts in the Comment Section.
I hope this information helps you.
Good Luck!