A bullet of mass 20 g and moving with 600 ms-1 collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision

Question

A bullet of mass 20 g and moving with 600 ms-1 collides with a block of mass 4 kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision (A) 200 ms-1 (B) 150 ms-1 (C) 400 ms-1 (D) 300 ms-1

Tardigrade Admin · Accepted Answer

According to conservation of linear momentum

m_{1} v_{1} = m_{1} v + m_{2} v_{2}

where

v_{1}

, is velocity of bullet before collision,

v

is velocity of bullet after the collision and

v_{2}

is the velocity of block.
=

0.02 \times 600 = 0.02 v + 4 v_{2}

Here

v_{2} = 2 g h = 2 \times 10 \times 0.2 = 2 m s^{- 1}

=

0.02 \times 600 = 0.02 v + 4 \times 2

\Rightarrow 0.02 v = 12 - 8

\Rightarrow v = \frac{4}{0.02} = 200 m s^{- 1}

Q. A bullet of mass $20 g$ and moving with $600 m s^{- 1}$ collides with a block of mass $4 k g$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height $0.2 m$ after collision

$200 m s^{- 1}$

$150 m s^{- 1}$

$400 m s^{- 1}$

$300 m s^{- 1}$

Q. A bullet of mass 20g and moving with 600ms−1 collides with a block of mass 4kg hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height 0.2m after collision

200ms−1

150ms−1

400ms−1

300ms−1

Q. A bullet of mass $20 g$ and moving with $600 m s^{- 1}$ collides with a block of mass $4 k g$ hanging with the string. What is velocity of bullet when it comes out of block, if block rises to height $0.2 m$ after collision

$200 m s^{- 1}$

$150 m s^{- 1}$

$400 m s^{- 1}$

$300 m s^{- 1}$