A bullet of mass 10 g moving horizontally with a velocity of 400 m s- 1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be

Question

A bullet of mass 10 g moving horizontally with a velocity of 400 m s- 1 strikes a wooden block of mass 2 kg which is suspended by a light inextensible string of length 5 m. As a result, the centre of gravity of the block is found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges out horizontally from the block will be (A) 100 m s- 1 (B) 80 m s- 1 (C) 120 m s- 1 (D) 160 m s- 1

Tardigrade Admin · Accepted Answer

CM rises through height h, so its velocity after collision =√2 g h Apply conservation of linear momentum. 0.01× 400=2× √2 g h+0.01× v ⇒ v=120 m/s

$100 m s^{- 1}$

$80 m s^{- 1}$

$120 m s^{- 1}$

$160 m s^{- 1}$

CM rises through height h, so its velocity after collision $= 2 g h$
Apply conservation of linear momentum.
$0.01 \times 400 = 2 \times 2 g h + 0.01 \times v$
$\Rightarrow v = 120 m / s$

100ms−1

80ms−1

120ms−1

160ms−1

CM rises through height h, so its velocity after collision =2gh​ Apply conservation of linear momentum. 0.01×400=2×2gh​+0.01×v ⇒v=120m/s

$100 m s^{- 1}$

$80 m s^{- 1}$

$120 m s^{- 1}$

$160 m s^{- 1}$

CM rises through height h, so its velocity after collision $= 2 g h$
Apply conservation of linear momentum.
$0.01 \times 400 = 2 \times 2 g h + 0.01 \times v$
$\Rightarrow v = 120 m / s$