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 m/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 velocity V of the bullet is [2011]

Question

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(4)

Let after $' t'$ time both ball and bullet hit the ground.

Then, $t = \sqrt{\frac{2 h}{g}} = \sqrt{\frac{2 \times 5}{10}} = 1 sec.$

After collision let $V_{ball}$ be velocity of ball and $V_{bullet}$ be velocity of bullet.

$So, 20 = V_{ball} \times 1 \Rightarrow V_{ball} = 20 m/s$

$100 = V_{bullet} \times 1 \Rightarrow V_{bullet} = 100 m/s$

By law of conservation of momentum

$0.01 V = 0.01 \times 100 + 0.2 \times 20$

$\Rightarrow 0.01 V = 1 + 4$

$\Rightarrow V = \frac{5}{0.01} = 500 m/s$

Solution

1 $250 m/s$

2 $250 \sqrt{2} m/s$

3 $400 m/s$

4 $500 m/s$

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Solution

1 250 m/s

2 2502 m/s

3 400 m/s

4 500 m/s

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1 $250 m/s$

2 $250 \sqrt{2} m/s$

3 $400 m/s$

4 $500 m/s$