• # question_answer Momentum of a copper sphere of radius r units is $x$ units, then momentum of another copper sphere of radius 2r units moving with the same velocity is $2x$ units. (T/F).

 Case-I Case-II ${{r}_{1}}=r\,\,units$ ${{r}_{2}}=2r\,\,units$ ${{P}_{1}}=x\,\,units$ ${{P}_{2}}=?$ ${{v}_{1}}=u\,\,units$ ${{v}_{2}}=u\,\,units$
$P=m\times v$ We know that mass $=m=V\times d$ $\Rightarrow m=d\times \frac{4}{3}\pi {{r}^{3}}$($\because$volume of sphere$=\frac{4}{3}\pi {{r}^{3}}$) $\therefore P=m\times v=\left( d\times \frac{4}{3}\pi {{r}^{3}} \right)\times u$ Both the spheres are made of same material, density is same in both the cases. Also their velocities are same $\Rightarrow P\propto {{r}^{3}}$ .......(1) Applying (1) to both the cases, we get ${{P}_{1}}\propto r_{1}^{3}$ ${{P}_{2}}\propto r_{2}^{3}$ $\Rightarrow x\propto {{r}^{3}}$ ............(2) $\Rightarrow {{P}_{2}}\propto {{(2r)}^{3}}\propto 8{{r}^{3}}$ .....(3) Dividing (2) by (3), we get $\frac{(2)}{(3)}=\frac{x}{{{p}_{2}}}=\frac{{{r}^{3}}}{8{{r}^{3}}}=\frac{x}{{{P}_{2}}}=\frac{1}{8}\Rightarrow {{P}_{2}}=8x$ Therefore, the final momentum is$2800\text{ }v=70,000\text{ }or\text{ }v=\frac{70000}{2800}$units. Therefore, the given statement is false.