Question

A particle of mass 100 g moving at an initial speed u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.2 J after the collision, what could be the maximum and the minimum value of u.

Answer 1

Final K.E.=0.2 J

Initial K.E.=$\frac{1}{2}m{v}_1^2+0$

=$\frac{1}{2}\times 0.1\times u^2=0.05u^2$

$m{v}_1+m{v}_2=mu$

where $v_1$ and $v_2$ are final velocities of 1st and 2nd block respectively

$\Rightarrow v_1+v_2=u...\left(i\right)$

$(v_1-v_2)+e(u_1-u_2)=0$

$\Rightarrow eu=v_2-v_1$

$[u_2=0,u_1=u]...\left(2\right)$

Adding equation (1) and (2),
$2v_2=(1+e)u$

$\Rightarrow v_2=\left(\frac{u}{2}\right)(1+e)$

$\therefore v_1=u-\frac{u}{2}(1+e)$

$v_1=\frac{u}{2}(1-e)$

$Given,\frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=0.2$

$\Rightarrow v_1^2+v_2^2=4$

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

$\Rightarrow u^2=\frac{8}{1+e^2}$

For maximum value of u,denominator should be minimum,
$\Rightarrow e=0\left(inelasticcollision\right)$

$\Rightarrow u^2=8$

$\Rightarrow u=2\sqrt{2}\frac{m}{s}$

For minimum value of u,denominator should be maximum ,
$\Rightarrow e=1\left(perfectelasticcollision\right)$

$\Rightarrow u^2=4$

$\Rightarrow u=2\frac{m}{s}$