Question

A particle of mass 100g moving at an initial spped u collides with another particle of same mass kept initially at rest. If the total kinetic energy becomes 0.5J after the collision, what could be the minimum and the maximum value of u.

• A. 2 m/s, 4 m/s
• B. 2 m/s, 2$\sqrt{2}$m/s
• C. $2\sqrt{2}$m/s , 2 m/s
• D. 2 m/s ,2 m/s

Answer 1

The correct option is B 2 m/s, 2$\sqrt{2}$m/s

Final K.E. = 0.2J
Initial K.E. $=\frac{1}{2}m{V}_1^2+0=\frac{1}{2}\times 0.1u^2=0.05u^2$
$m{v}_1=m{v}_2^{\prime }=mu$
Where $v_{1}and{v}_2$ are final velocities of $1^{st}and{2}^{nd}$ block respectively.
$\Rightarrow v_1+v_2=u$ . . . (1)
$(v_1-v_2)+t(a_1-u_2)=0\Rightarrow ta=v_2-v_1$ . . . (2)
$u_2=0,u_1=u$
Adding Eq.(1) and Eq.(2)
$2v_2=(1+l)u\Rightarrow v_2=\left(\frac{u}{2}\right)(1+l)$
$\therefore v_1=u-\frac{u}{2}-\frac{u}{2}l$
$v_1=\frac{u}{2}(1-l)$
$Given\left(\frac{1}{2}\right)m{v}_1^2+\left(\frac{1}{2}\right)m{v}_2^2=0.2$
$\Rightarrow v_1^2+v_2^2=4$
$\Rightarrow \frac{u^2}{4}(1-l{)}^2+\frac{U^2}{4}(1+l{)}^2=4\Rightarrow \frac{u^2}{2}(1+l^2)=4\Rightarrow u^2=\frac{8}{1+l^2}$
For maximum value of u, denominator should be minimum,
$\Rightarrow l=0$
$\Rightarrow u^2=8\Rightarrow u=2\sqrt{2}m/s$
For minimum value of u, denominator should be maximum,
$\Rightarrow l=1$
$u^2=4\Rightarrow u=2m/s$