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

  • Option 1)

    100 m s-1

  • Option 2)

    80 m s-1

  • Option 3)

    120 m s-1

  • Option 4)

    160 m s-1

 

Answers (1)

best_answer

As we learnt in

Impulse Momentum Theorem -

vec{F}=frac{dvec{p}}{dt}

int_{t_{1}}^{t_{2}}vec{F}dt=int_{p_{1}}^{p_{2}}vec{dp}

- wherein

If igtriangleup t  is increased, average force is decreased

 Apply conservation of momentum

\frac{10}{1000}\times400+0=2V_{1}+\frac{10}{1000}\times V_{2}

4=2V_{1}+0.01V_{2}....(i)

W.E.T\Rightarrow W=\Delta KE

2\times 10\times0.1=\frac{1}{2}\times 2\times V_{1}^{2}

V_{1}=\sqrt{2}=1.4 m/s \rightarrow put\ in (1)

4=2\times 1.4+0.01V_{2}
V_{2}=120 m/s

 


Option 1)

100 m s-1

Incorrect

Option 2)

80 m s-1

Incorrect

Option 3)

120 m s-1

Correct

Option 4)

160 m s-1

Incorrect

Posted by

divya.saini

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