Question

$A$ ball of mass 0.2 kg rests on a vertical post of height 5 m. A bullet of mass 0.01 kg, traveling with a velocity $Vm/s$ in a horizontal direction, hits the centre of the ball. $A$fter the collision, the ball and bullet travel independently. The ball hits the ground at a distance of 20 $m$ and the bullet at a distance of 100 $m$ from the foot of the post. The initial velocity $V$ of the bullet is

• A. 250 $m/s$
• B. $250\sqrt{2}m/s$
• C. 400 $m/s$
• D. 500 $m/s$

Answer 1

Answer: (D)

500 $m/s$

Let the mass of bullet be m and mass of ball be M
initially ball is at height 5m and t rest , the only acceleration is due to gravity
therefore applying equation of motion

$S=ut+\frac{1}{2}g{t}^2$

$5=0+\frac{1}{2}\left(10\right)t^2$

therefore t $=$ 1 sec

so $V\left(ball\right)=20m/sec$

$V\left(bullet\right)$ = $100m/sec$

So by collision
$M\times V(ball:final)+m\times V(bullet:final)$= $M\times V(boll:initial)+m\times V(bullet:initial)$

$0.01\times V\left(bulletfinal\right)=0.01\times 100+0.20\times 20$

$V=500m/sec$