Question

A block of mass m 2.0 kg moving at $2.0\frac{m}{s}$ collides head on with another block of equal mass kept at rest.(a) Find the maximum possible loss in kinetic energy due to the collision.(b) If the actual loss in kinetic energy is half of this maximum,find the coefficient of restitution.

Answer 1

(a) Mass of block=2 kg

and speed=$2\frac{m}{s}$

Mass of 2nd block=2 kg

Let final velocity of 2nd block=v

Using law of conservation of momentum,$2\times 2=(2+2)-v$

$\Rightarrow v=1\frac{m}{s}$

$\therefore$ Loss in K.E. in elastic collision

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

$=4-2=2J$

(b) Actual loss=$\frac{Maximumloss}{2}=1J$

$\left(\frac{1}{2}\right)\times 2\times (2{)}^2-\left(\frac{1}{2}\right)2\times v_1^2-\left(\frac{1}{2}\right)\times 2v_2^2=1$

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

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

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

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

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

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