As we have mentioned in the previous notes that in inelastic collision only momentum conservation holds and not the conservation of the kinetic energy. Infact in real world collisions are generally inelastic. Here we'll discuss the

We'll find an expression for the fraction of kinetic energy loss due to

$$v=(\frac{m_{1}}{m_{1}+m_{2}})u_{1}+(\frac{m_{2}}{m_{1}+m_{2}})u_{2}....(2)$$

so the kinetic energy of the stuck ball is just simply,

If $u_{2}=0$ then the initial kinetic energy is simply,

$$\Delta K.E=\frac{1}{2}m_{1}u_{1}^{2}\left \{1- \left ( \frac{m_{1}}{m_{1}+m_{2}} \right )\right \}.....(6)$$

Hence the

$$=\left \{1- \left ( \frac{m_{1}}{m_{1}+m_{2}} \right )\right \}$$

$$\frac{\Delta K.E}{Initial K.E}=\frac{m2}{m_{1}+m_{2}}.....(7)$$

Some of the

Relevant Exercise:

**change**in*Kinetic energy*due to inelastic collision, in which objects stick together after the collision or in other words the perfect inelastic collision.We'll find an expression for the fraction of kinetic energy loss due to

**perfect inelastic collision.**

**As we know that the expression for the final velocity in case of the inelastic collision has already been derived in the previous notes in the exercise problem**

**,**the expression that we obtained was,
$m_{1}u_{1}+m_{2}u_{2}=m_{1}v+m_{2}v......(1)$

Rearranging the terms gives us,$$v=(\frac{m_{1}}{m_{1}+m_{2}})u_{1}+(\frac{m_{2}}{m_{1}+m_{2}})u_{2}....(2)$$

so the kinetic energy of the stuck ball is just simply,

$$\frac{1}{2}(m_{1}+m_{2})\left \{ \left ( \frac{m_{1}}{m_{1}+m_{2}}u_{1}+\frac{m_{1}}{m_{1}+m_{2}}u_{2} \right )^{2} \right \}.....(3)$$

for simplicity assume, $u_{2}=0$,
$$\frac{1}{2}(m_{1}+m_{2})\left \{ \left (
\frac{m_{1}}{m_{1}+m_{2}}u_{1}\right
)^{2} \right \}.....(4)$$

If $u_{2}=0$ then the initial kinetic energy is simply,

$\frac{1}{2}m_{1}u_{1}^{2}.....(5)$

The difference in the kinetic energy is then,$$\Delta K.E=\frac{1}{2}m_{1}u_{1}^{2}\left \{1- \left ( \frac{m_{1}}{m_{1}+m_{2}} \right )\right \}.....(6)$$

Hence the

**fractional change in kinetic energy**(i.e. $\Delta K.E/\mathrm{initial (K.E)}$)is,$$=\left \{1- \left ( \frac{m_{1}}{m_{1}+m_{2}} \right )\right \}$$

$$\frac{\Delta K.E}{Initial K.E}=\frac{m2}{m_{1}+m_{2}}.....(7)$$

**Exercise: Play with this fractional kinetic energy calculator.**Some of the

**example**of the inelastic collision can be found here, in the video example.Relevant Exercise:

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