A block of mass $2.0 \mathrm{~kg}$ moving at $2.0 \mathrm{~m} / \mathrm{s}$ 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 co-efficient of restitution.
(a) Use C.O.L.M
$2 .(2)+0=2\left(v_{1}+v_{2}\right)$
$v_{1}+v_{2}=2$
$(K . E)_{\text {loss }}=\frac{1}{2} m\left(v_{1}^{2}+v_{2}^{2}\right)-\frac{1}{2} m(4)$
For maximizing, take derivative.
$(K . E)_{\text {loss }}=\frac{1}{2} m\left(\left(2-v_{2}\right)^{2}+v_{2}^{2}\right)-\frac{1}{2} \cdot 4 . m$
$\frac{d(K . E)_{\text {loss }}}{d v_{2}}=\frac{1}{2} m\left(2\left(2-v_{2}\right)+2 v_{2}\right)=0$
$\Rightarrow\left(2-v_{2}\right)=-v_{2}$
$\Rightarrow v_{2}=1_{\mathrm{m} / \mathrm{s}}$
$(K . E)_{\left.l o s s\right|_{v_{h}=0}}=\left|\frac{1}{2} .2\left((1)^{2}+(1)^{2}\right)-\frac{1}{2} \times 2 \times 4\right|$
$=|2-4|=2_{\mathrm{J}}$
$(\mathrm{b})(K \cdot E)_{\text {loss }}=1$
$v_{1}^{2}+v_{2}^{2}-4=-1$
$v_{1}^{2}+v_{2}^{2}=3, \quad v_{1}+v_{2}=2$
$v_{1}^{2}+v_{2}^{2}+2 v_{1} v_{2}=4$
$v_{1} v_{2}=\frac{4-3}{2}=+\frac{1}{2}$
$\left(v_{1}-v_{2}\right)^{2}=v_{1}^{2}+v_{2}^{2}-2 v_{1} v_{2}=3-1=2$
$\left(v_{1}-v_{2}\right)=\sqrt{2}$
$e=\frac{v_{1}-v_{2}}{v}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.