A ball of mass 0.2 kg rests on a vertical post of height 5 m. A b… - Vidyadip

MCQ

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 $V / s$ in a horizontal direction, hits the centre of the ball. After 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
$400 \ m / s$
✓
$500 \ m / s$

✓

Answer

Correct option: D.

$500 \ m / s$

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

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

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

therefore $t=1 \ sec$

$\text { so } V ( ball )=20 \ m / sec$

$V \text { (bullet) }=100 \ m / sec$

So by collision

$M \times V \text { (ball : final })+ m \times V \text { (bullet : final })=$

$M \times V \text { (boll : initial })+ m \times V \text { (bullet : initial })$

$0.01 \times V \text { (bulletfinal) }=0.01 \times 100+0.20 \times 20$

$V=500 \ m / sec$

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