A bullet of mass 10 g moving horizontally with a velocity of 400 ms-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 ms-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 ms-1 (B) 80 ms-1 (C) 120 ms-1 (D) 160 ms-1

Tardigrade Admin · Accepted Answer

Velocity of block = √2gh = √2 × g × 0.1 = √2 m/s Pi = Pf (10 × 10-3 ) × 400 + 0 = 2 × √2 + (10 × 10-3 ) V V = ((4 - 2 √2)/10 × 10-3) = 120 m/s

$100 m s^{- 1}$

$80 m s^{- 1}$

$120 m s^{- 1}$

$160 m s^{- 1}$

Velocity of block = $2 g h = 2 \times g \times 0.1 = 2 m / s$
$P_{i} = P_{f}$
$(10 \times 1 0^{- 3}) \times 400 + 0 = 2 \times 2 + (10 \times 1 0^{- 3}) V$
$V = \frac{( 4 - 2 2 )}{10 \times 1 0 ^{- 3}} = 120 m / s$

100ms−1

80ms−1

120ms−1

160ms−1

Velocity of block = 2gh​=2×g×0.1​=2​m/s Pi​=Pf​ (10×10−3)×400+0=2×2​+(10×10−3)V V=10×10−3(4−22​)​=120m/s

$100 m s^{- 1}$

$80 m s^{- 1}$

$120 m s^{- 1}$

$160 m s^{- 1}$

Velocity of block = $2 g h = 2 \times g \times 0.1 = 2 m / s$
$P_{i} = P_{f}$
$(10 \times 1 0^{- 3}) \times 400 + 0 = 2 \times 2 + (10 \times 1 0^{- 3}) V$
$V = \frac{( 4 - 2 2 )}{10 \times 1 0 ^{- 3}} = 120 m / s$