In this section we'll discuss or in specific derive the simple fact that

If you still have doubts about the concepts of the elastic collision and inelastic collision click here.

Now consider the following figure,

To show, Relative speed of approach= Relative speed of separation i.e,

$$U_{1}-U_{2}=V_{2}-V_{1}$$

From the conservation of Kinetic energy and Momentum we have,

$$m_{1}U_{1}+m_{2}U_{2}=m_{1}V_{1}+m_{2}V_{2} ........(1)$$

$$\frac{1}{2}m_{1}U_{1}^{2}+\frac{1}{2}m_{2}U_{2}^{2}=\frac{1}{2}m_{1}V_{1}^{2}+\frac{1}{2}m_{2}V_{2}^{2}........(2)$$

The

Suppose we arrange equation $(1)$ and equation $(2)$ as follows,

$$m_{1}(U_{1}-V_{1})=m_{2}(V_{2}-U_{2}) ........(3)$$

$$m_{1}(U_{1}^{2}-V_{1}^{2})=m_{2}(V_{2}^{2}-U_{2}^{2})........(4)$$

Now then substituting $m_{1}/m_{2}$ from eqn $(3)$ to eqn $(4)$ we finally have,

$$(U_{1}+V_{1})=(V_{2}+U_{2})........(5)$$

We have by rearranging the term again,

$$U_{1}-U_{2}=V_{2}-V_{1}....(6)$$

Which is the required expression we had to show, Thus

*Well If you have heard of coefficient of restitution, it would make easy your proving. The coefficient of restitution is simply,

$e$ =

You can thus easily see that for the $e=1$ the relation $(5)$ does hold true. That's all we'll discuss about restitution.

**"in perfectly elastic collision the relative speed of approach is equal to the relative speed of separation."**If you still have doubts about the concepts of the elastic collision and inelastic collision click here.

Now consider the following figure,

To show, Relative speed of approach= Relative speed of separation i.e,

$$U_{1}-U_{2}=V_{2}-V_{1}$$

**Proof:**From the conservation of Kinetic energy and Momentum we have,

$$m_{1}U_{1}+m_{2}U_{2}=m_{1}V_{1}+m_{2}V_{2} ........(1)$$

$$\frac{1}{2}m_{1}U_{1}^{2}+\frac{1}{2}m_{2}U_{2}^{2}=\frac{1}{2}m_{1}V_{1}^{2}+\frac{1}{2}m_{2}V_{2}^{2}........(2)$$

The

**trick**is in the rearranging the equations,Suppose we arrange equation $(1)$ and equation $(2)$ as follows,

$$m_{1}(U_{1}-V_{1})=m_{2}(V_{2}-U_{2}) ........(3)$$

$$m_{1}(U_{1}^{2}-V_{1}^{2})=m_{2}(V_{2}^{2}-U_{2}^{2})........(4)$$

Now then substituting $m_{1}/m_{2}$ from eqn $(3)$ to eqn $(4)$ we finally have,

$$(U_{1}+V_{1})=(V_{2}+U_{2})........(5)$$

We have by rearranging the term again,

$$U_{1}-U_{2}=V_{2}-V_{1}....(6)$$

Which is the required expression we had to show, Thus

**"in perfectly elastic collision the relative speed of approach is equal to the relative speed of separation."***Well If you have heard of coefficient of restitution, it would make easy your proving. The coefficient of restitution is simply,

$e$ =

**velocity of separation / velocity of approach**You can thus easily see that for the $e=1$ the relation $(5)$ does hold true. That's all we'll discuss about restitution.

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