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 minimum and the maximum value of u.

Answer 1

It is given that:
Mass of particles = 100 g
Initial speed of the first particle = u
Final K.E. of the system after collision = 0.2J



Initial K.E. of the system, before collision = $\frac{1}{2}m{u}^2+0$
i.e. Initial K.E. = $\frac{1}{2}\times 0.1\times u^2=0.05u^2$

Let v₁ and v₂ be the final velocities of the first and second block respectively.

By law of conservation of momentum, we know:

$m{v}_1+m{v}_2=mu$
$$\begin{gathered} \Rightarrow v_1+v_2=u...\left(1\right) \\ (v_1-v_2)+e(u_1-u_2)=0 \\ \Rightarrow eu=v_2-v_1...\left(2\right)[\mathrm{Putting}{u}_2=0,u_1=u] \\ \mathrm{Adding}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}: \\ 2v_2=(1+e)\mathrm{u} \\ \Rightarrow {\mathrm{v}}_2=\left(\frac{u}{2}\right)(1+e) \\ \therefore {\mathrm{v}}_1=u-\frac{\mathrm{u}}{2}(1+e) \\ {\mathrm{v}}_1=\frac{\mathrm{u}}{2}(1-e) \\ \mathrm{According}\mathrm{to}\mathrm{given}\mathrm{condition}, \\ \frac{1}{2}m{v}_1^2+\frac{1}{2}m{v}_2^2=0.2 \\ \Rightarrow v_1^2+v_2^2=4 \\ \Rightarrow \frac{u^2}{2}\left(1+e^2\right)=4 \\ \Rightarrow u^2=\frac{8}{1+e^2} \end{gathered}$$

For maximum value of u, denominator should be minimum in the above equation.
i.e. e = 0
⇒ u² = 8

$\Rightarrow u=2\sqrt{2}\mathrm{m}/\mathrm{s}$

For minimum value of u, denominator should have maximum value.
i.e. e = 1
⇒ u² = 4

⇒ u = 2 m/s