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  • Please help! A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

 Option 2)

0.16\; J
Option 3)

1.00\; J

 Option 4)

0.67\; J

Answers (3)

best_answer

As we discussed in

Perfectly Elastic Collision -

Law of conservation of momentum and that of Kinetic Energy hold good.

- wherein

\frac{1}{2}m_{1}u_{1}^{2}+\frac{1}{2}m_{2}u_{2}^{2}= \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}v_{2}

m_{1},m_{2}:masses

u_{1},v_{1}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{1}

u_{2},v_{2}:initial \: and\: final \: velocity\: of \: the\: mass\ m_{2}

 

 Initial kinetic energy =\frac{1}{2}\times(0.50)\times4=1J

After collision momentum = 1.5 v

From momentum conservation 

    1.5 v = 1

\Rightarrow\ v=\frac{2}{3}m/s

Final kinetic energy =\frac{1}{2}\times(1.5)\times(\frac{2}{3})^{2}=\frac{3}{4}\times\frac{4}{9}-\frac{1}{3}J

Loss in kinetic energy =K.E_{f}-K.E_{i}= -\frac{2}{3}J


Option 1)

0.34\; J

This is an incorrect option.

Option 2)

0.16\; J

This is an incorrect option.

Option 3)

1.00\; J

This is an incorrect option.

Option 4)

0.67\; J

This is the correct option.

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