Question

A block of mass $2kg$ moving at $2.0m/s$ collides head on with another block of equal mass kept at rest. If the loss in kinetic energy of the system is half of the maximum possible loss of kinetic energy of system, if the coefficient of restitution $\frac{\lambda }{\sqrt{2}}$. Then find $\lambda$.

Answer 1

Mass of first block=mass of second block=$2$kg.

Velocity of first=$2m{s}^{-1}$

For maximum energy both block move same velocity($v$)

By momentum conservation,

$2\times 2=(2+2)v$

$v=1m{s}^{-1}$

Loss of energy=$\frac{1}{2}\times 2(2{)}^2-\frac{1}{2}\times 4\times 1^2$

=4-2=2J

If losses is half maximum loss second case losses=1J

By momentum conservation,$2\times 2=2\times v_1+2\times v_2$

$v_1+v_2=2$ where $v_1$ & $v_2$ are velocities after collision respectively.

$e=\frac{v_2-v_1}{v}$

$v_2-v_1=ev$

Now, {$\frac{1}{2}\times 2(2{)}^2$}-{$\frac{1}{2}\times 2v_1^2\times 2v_2^2$}=1

$4-(v_1^2+v_2^2)$=1

$4-\frac{(1+e^2)\times 4}{2}$=1

$2(1+e^2)$=3

$1+e^2=\frac{3}{2}$

$e^2=\frac{1}{2}$

$e=\frac{1}{\sqrt{2}}$

Hnece $\lambda =1$