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An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 ms^{-1} and 2 kg second part moving with a velocity of 8 ms^{-1}. If the third part flies off with a velocity of 4 ms^{-1}, its mass would be

  • Option 1)

    7 kg

  • Option 2)

    17 kg

  • Option 3)

    3 kg

  • Option 4)

    5 kg

 

Answers (1)

best_answer

As we learnt  in

Inelastic Collision -

Law of conservation of momentum hold good but that of kinetic energy is not

- wherein

frac{1}{2}m_{1}u_{1}^{2}+frac{1}{2}m_{2}u_{2}^{2} eq frac{1}{2}m_{1}v_{1}^{2}+frac{1}{2}m_{2}v_{2}^{2}

m_{1}u_{1}+m_{2}u_{2}= m_{1}v_{1}+m_{2}

m_{1},m_{2}: masses

u_{1},v_{1}: initional : and : final: velocities: of : mass : m_{1}

u_{2},v_{2}: initional : and : final: velocities: of : mass : m_{2}

 

 Net momentum of the two pieces 1kg and 2kg

P = \sqrt{12^{2}+16^{2}} = 20 kg ms^{-1}

\therefore momentum of third piece = 20 kg ms^{-1}

V = 4 ms^{-1}

\therefore mass of 3rd Piece = \frac{20}{4} = 5 kg 


Option 1)

7 kg

This solution is incorrect

Option 2)

17 kg

This solution is incorrect

Option 3)

3 kg

This solution is incorrect

Option 4)

5 kg

This solution is correct

Posted by

prateek

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