question_answer

The velocity of a iron sphere of radius r units V units. What is the velocity of another iron sphere of radius 1r units, with the same momentum as that of the first one?

Answer:

Let the momentum of the two gold spheres be\[{{P}_{1}}={{P}_{2}}=P\]units respectively.

Case-I	Case-II
\[{{r}_{1}}=r\,\,units.\]	\[{{r}_{2}}=2r\,\,units.\]
\[{{V}_{1}}=V\,\,units\]	\[{{V}_{2}}=?\]
\[P=m\times V\]

But \[m=d\times v=d\times \frac{4}{3}\pi {{r}^{3}}\] \[\left( \because volume\,\,of\,\,sphere=\frac{4}{3}\pi {{r}^{3}} \right)\] \[\Rightarrow P=\left( d\times \frac{4}{3}\pi {{r}^{3}} \right)\times V\]                  …….. (1) As same material is taken, its density is same. \[\Rightarrow \,P\propto {{r}^{3}}\times V\Rightarrow P=k({{r}^{3}}V)\]              ……. (2) Applying (2) to both the cases, we get, \[{{P}_{1}}=k[{{({{r}_{1}})}^{3}}\times {{V}_{1}}]\,{{P}_{2}}=k[{{({{r}_{2}})}^{3}}\times {{V}_{2}}]\] \[\Rightarrow \,\,{{P}_{1}}=k[{{(r)}^{3}}\times V]\]                         ……. (3) \[{{P}_{2}}=k[{{(2r)}^{3}}\times {{V}_{2}}]\] \[\Rightarrow \,\,{{P}_{2}}=k(8{{r}^{3}}\times {{V}_{2}})\]                           ……. (4) Dividing equation (3) by (4), we get, \[\therefore \frac{(3)}{(4)}=\frac{{{P}_{1}}}{{{P}_{2}}}=\frac{k({{r}^{3}}\times V)}{K(8{{r}^{3}}\times {{V}_{2}})}=\frac{V}{8{{V}_{2}}}\] \[\Rightarrow \frac{P}{P}=\frac{V}{8{{V}_{2}}}\Rightarrow 1=\frac{V}{8{{V}_{2}}}\Rightarrow 8{{P}_{2}}=V\] \[\Rightarrow {{V}_{2}}=\frac{V}{8}\] \[\therefore \]velocity of second gold sphere of radius 2r units, with same momentum as that of the first one is \[\frac{1}{8}\]times velocity of the first gold sphere.