First slide

The steps in collision

Question

A bullet of mass 20 g moving with $600 {ms}^{- 1}$ collides with a block of mass 4 kg hanging with the string of length 0.4 m. What is velocity of bullet when it comes out of block, if block rises to height 0.2 m after collision? (Take, $g = 10 {ms}^{- 2}$ )

Moderate

A

$200 {ms}^{- 1}$
B

$150 {ms}^{- 1}$
C

$400 {ms}^{- 1}$
D

$300 {ms}^{- 1}$

Solution

Velocity of block just after collision $= \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.2}$
$= 2 m s^{- 1}$
Now, applying conservation of linear momentum just before and just after collision,
0.02 $\times$ 600 = 4 $\times$ 2 + 0.02 $\times$ v
$∴$ $v = 200 {ms}^{- 1}$

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