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Standard XII

Physics

Momentum

A bullet of m...

Question

A bullet of mass $10$ g moving horizontally at a speed of $140$ m $/$ s strikes a block of mass $100$ g attached to a string like a simple pendulum. The bullet penetrates the block and emerges then on the other side. If the block rises by $80$ cm, then find the final velocity of bullet.

80 m / s

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100 m / s

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120 m / s

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140 m / s

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Solution

The correct option is B $100 m / s$
Let the speed of the bullet be $v_{1}$ . Let the common velocity of the bullet and the block, after the bullet is attached like a simple pendulum is $v_{2}$ .

By the principle of conservation of linear momentum

$\frac{10}{1000} \times 140 + \frac{100}{1000} \times 0$

$= \frac{100}{1000} v_{1} + \frac{10}{1000} v_{2}$

$140 = 10 v_{1} + v_{2}$ ..........(i)

For the block, $0 = v_{1}^{2} - 2 g (\frac{80}{100})$

$\Rightarrow v_{1} = 4 m / s$

$\Rightarrow 140 = 10 \times 4 + v_{2}$ [from Eq. (i)]

$\Rightarrow v_{2} = 100 m / s$

Final velocity of bullet $= 100 m / s$ .

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A bullet of mass 10g moving horizontally at a speed of 140m/s strikes a block of mass 100g attached to a string like a simple pendulum. The bullet penetrates the block and emerges then on the other side. If the block rises by 80cm, then find the final velocity of bullet.

A bullet of mass $10$ g moving horizontally at a speed of $140$ m $/$ s strikes a block of mass $100$ g attached to a string like a simple pendulum. The bullet penetrates the block and emerges then on the other side. If the block rises by $80$ cm, then find the final velocity of bullet.