Physics - Center of Mass Question with Solution | TestHub

Velocity of first block before collision, $v_1^2=1^2-2(2)\times 0.16$ $=1-0.64$

$${\mathrm{v}}_1=0.6\mathrm{m}/\mathrm{s}$$

By conservation of momentum, $2\times 0.6=2{\mathrm{v}}_1^{\prime }+4{\mathrm{v}}_2^{\prime }$ Also ${\mathrm{v}}_2^{\prime }-{\mathrm{v}}_1^{\prime }={\mathrm{v}}_1$ for elastic collision It gives ${\mathrm{v}}_2^{\prime }=0.4\mathrm{m}/\mathrm{s}$

$${\mathrm{v}}_1^{\prime }=-0.2\mathrm{m}/\mathrm{s}$$

Now distance moved after collision

$$\begin{array}{rlr}& {\mathrm{s}}_1=\frac{(0.4{)}^2}{2\times 2}\&{\mathrm{s}}_2=\frac{(0.2{)}^2}{2\times 2}& \therefore \mathrm{s}={\mathrm{s}}_1+{\mathrm{s}}_2=0.05\mathrm{m}=5\mathrm{cm}\end{array}$$