# Q. A light particle moving horizontally with a speed of 12 m/s strikes a very heavy block moving in the same direction at 10 m/s.The collision is one-dimensional and elastic. After the...

Q. A light particle moving horizontally with a speed of 12 m/s strikes a very heavy block moving in the same direction at 10 m/s.The collision is one-dimensional and elastic. After the collision,the particle will

A) move at 2m/s in its original direction

B) move at 8m/s in its original direction

C) move at 8m/s opposite to its original direction

D) move at 12 m/s opposite to its original direction

### 1 Answer | Add Yours

In an elastic collision the kinetic energy is conserved.

When two particles moving along the same axis take part in such collision,

Conservation of momentum requires:

`m_1v_1+m_2v_2=m_1v'_1+m_2v'_2`

`rArr m_1(v_1-v'_1)=m_2(v_2-v'_2)` --- (i)

Conservation of kinetic energy requires:

`1/2(m_1v_1^2+m_2v_2^2)=1/2(m_1v'_1^2+m_2v'_2^2)`

`rArr m_1(v_1^2-v'_1^2)=m_2(v_2^2-v'_2^2)`

`rArr m_1(v_1-v'_1)(v_1+v'_1)=m_2(v_2-v'_2)(v_2+v'_2)` --- (ii)

Dividing by (i) yields:

`v_1+v'_1=v_2+v'_2`

Assuming the heavier block to be so heavy that its velocity remain unperturbed after the collision with the light patrticle, the final velocity of the light particle (v'_1) can be obtained as:

`12+v'_1=10+10`

`rArr v'_1=8` m/s

So, the light particle will move with a velocity of 8 m/s in tha same direction (option B).

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