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 $g = 10\,ms^{-1}$ )

Vedclass pdf generator app on play store
Vedclass iOS app on app store

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, $m_{1} v=m_{2} u$ $\therefore v=\frac{m_{2} u}{m_{1}}=\frac{1}{100} \times 40=0.4 \mathrm{~m} / \mathrm{s}$

886-s186

Similar Questions

A body is moving with a velocity $v$, breaks up into two equal parts. One of the part retraces back with velocity $v$. Then the velocity of the other part is

A bullet mass $10\, gm$ is fired from a gun of mass $1\,kg$. If the recoil velocity is $5\, m/s$, the velocity of the muzzle is ........ $m/s$

You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface

A bullet of mass $5 \,g$ is shot from a gun of mass $5 \,kg$. The muzzle velocity of the bullet is $500\, m/s$. The recoil velocity of the gun is ........... $m/s$

A projectile is fired with velocity $u$ at an angle $\theta$ with horizontal. At the highest point of its trajectory it splits up into three segments of masses $m, m$ and $2 \,m$. First part falls vertically downward with zero initial velocity and second part returns via same path to the point of projection. The velocity of third part of mass $2 \,m$ just after explosion will be