A moving mass of 8 kg collides elastically with a stationary mass of 2 kg. If E be the initial kinetic energy of the mass, the kinetic energy left with it after collision will be

The correct option is 0.36 E.

Step-by-step derivation:

Let the mass of the moving object be $m_1=8\text{kg}$ and its initial velocity be $u_1$.
Let the mass of the stationary object be $m_2=2\text{kg}$ and its initial velocity be $u_2=0$.

The initial kinetic energy ($E$) of the moving mass is:
$E=\frac{1}{2}{m}_1{u}_1^2$

For a perfectly elastic collision, the final velocity ($v_1$) of the first mass after the collision is given by the formula:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1+\left(\frac{2m_2}{m_1+m_2}\right)u_2$

Since the second mass is initially stationary ($u_2=0$), the second term becomes zero:
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u_1$

Substitute the given masses $m_1=8\text{kg}$ and $m_2=2\text{kg}$ into the equation:
$v_1=\left(\frac{8-2}{8+2}\right)u_1$
$v_1=\left(\frac{6}{10}\right)u_1=0.6u_1$

The final kinetic energy ($E_f$) remaining with the first mass after the collision is:
$E_f=\frac{1}{2}{m}_1{v}_1^2$

Substituting $v_1=0.6u_1$ into the kinetic energy equation:
$E_f=\frac{1}{2}{m}_1{\left(0.6u_1\right)}^2$
$E_f=0.36·\left(\frac{1}{2}{m}_1{u}_1^2\right)$
$E_f=0.36E$

Therefore, the kinetic energy left with the moving mass after the collision is $0.36E$.