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).

Answer:

Let the velocities of both the spheres be u units

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.