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.

Open in App
Solution

## 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 = $\frac{1}{2}m{u}^{2}+0\phantom{\rule{0ex}{0ex}}$ i.e. Initial K.E. = $\frac{1}{2}×0.1×{u}^{2}=0.05{u}^{2}$ Let v1 and v2 be the final velocities of the first and second block respectively. By law of conservation of momentum, we know: $m{v}_{1}+m{v}_{2}=mu$ $⇒{v}_{1}+{v}_{2}=u...\left(1\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left({v}_{1}-{v}_{2}\right)+e\left({u}_{1}-{u}_{2}\right)=0\phantom{\rule{0ex}{0ex}}⇒eu={v}_{2}-{v}_{1}...\left(2\right)\left[\mathrm{Putting}{u}_{2}=0,{u}_{1}=u\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{the}\mathrm{equations}\left(1\right)\mathrm{and}\left(2\right),\mathrm{we}\mathrm{get}:\phantom{\rule{0ex}{0ex}}2{v}_{2}=\left(1+e\right)\mathrm{u}\phantom{\rule{0ex}{0ex}}⇒{\mathrm{v}}_{2}=\left(\frac{u}{2}\right)\left(1+e\right)\phantom{\rule{0ex}{0ex}}\therefore {\mathrm{v}}_{1}=u-\frac{\mathrm{u}}{2}\left(1+e\right)\phantom{\rule{0ex}{0ex}}{\mathrm{v}}_{1}=\frac{\mathrm{u}}{2}\left(1-e\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{According}\mathrm{to}\mathrm{given}\mathrm{condition},\phantom{\rule{0ex}{0ex}}\frac{1}{2}m{v}_{1}^{2}+\frac{1}{2}m{v}_{2}^{2}=0.2\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}⇒{v}_{1}^{2}+{v}_{2}^{2}=4\phantom{\rule{0ex}{0ex}}⇒\frac{{u}^{2}}{2}\left(1+{e}^{2}\right)=4\phantom{\rule{0ex}{0ex}}⇒{u}^{2}=\frac{8}{1+{e}^{2}}$ For maximum value of u, denominator should be minimum in the above equation. i.e. e = 0 ⇒ u2 = 8 $⇒u=2\sqrt{2}\mathrm{m}/\mathrm{s}$ For minimum value of u, denominator should have maximum value. i.e. e = 1 ⇒ u2 = 4 ⇒ u = 2 m/s

Suggest Corrections
0
Related Videos
Momentum Returns
PHYSICS
Watch in App
Explore more