Question

Body $\mathrm{A}$ of mass $4\mathrm{m}$ moving with speed $\mathrm{u}$ collides with another body $\mathrm{B}$ of mass $2\mathrm{m}$, at rest. The collision is head-on and elastic in nature. After the collision the fraction of energy lost by the colliding body $\mathrm{A}$ is

• A. $\frac{1}{9}$
• B. $\frac{8}{9}$
• C. $\frac{4}{9}$
• D. $\frac{5}{9}$

Answer 1

The correct option is B

$\frac{8}{9}$

Step 1: Given Data

Body $\mathrm{A}$ of mass $\left({\mathrm{m}}_1\right)=4\mathrm{m}$

Body $\mathrm{B}$ of mass $\left({\mathrm{m}}_2\right)=2\mathrm{m}$

Step 2: Formula used

Using the conservation equations, we can substitute the velocities of particles in terms of the initial velocity of neutrons.

The fractional loss in energy is the loss in kinetic energy divided by initial kinetic energy.

$\frac{∆\mathrm{KE}}{\mathrm{KE}}=\frac{4\times \left({\mathrm{m}}_1{\mathrm{m}}_2\right)}{{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}^2}$

Step 3: Determine the fraction of energy lost by the colliding body

Fractional loss of KE of colliding body,

$\frac{∆\mathrm{KE}}{\mathrm{KE}}=\frac{4\times \left({\mathrm{m}}_1{\mathrm{m}}_2\right)}{{\left({\mathrm{m}}_1+{\mathrm{m}}_2\right)}^2}$

By substituting the values, we get

$$\begin{gathered} =\frac{4\times \left(4\mathrm{m}\right)\times 2\mathrm{m}}{{\left(4\mathrm{m}+2\mathrm{m}\right)}^2} \\ =\frac{32{\mathrm{m}}^2}{36{\mathrm{m}}^2} \\ =\frac{8}{9} \end{gathered}$$

Hence, the fraction of energy lost is $\frac{8}{9}$, so option(b) is the correct answer.