# 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) Option 2) Option 3) Option 4)

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)

This is an incorrect option.

Option 2)

This is an incorrect option.

Option 3)

This is an incorrect option.

Option 4)

This is the correct option.

### Preparation Products

##### Knockout JEE Main Sept 2020

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 12999/- ₹ 6999/-
##### Rank Booster JEE Main 2020

This course will help student to be better prepared and study in the right direction for JEE Main..

₹ 9999/- ₹ 4999/-
##### Test Series JEE Main Sept 2020

Take chapter-wise, subject-wise and Complete syllabus mock tests and get in depth analysis of your test..

₹ 4999/- ₹ 1999/-
##### Knockout JEE Main April 2021

An exhaustive E-learning program for the complete preparation of JEE Main..

₹ 17999/- ₹ 11999/-