Answer:
Let the momentum of the two gold spheres be\[{{P}_{1}}={{P}_{2}}=P\]units respectively.
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.
Case-I
Case-II
\[{{r}_{1}}=r\,\,units.\]
\[{{r}_{2}}=2r\,\,units.\]
\[{{V}_{1}}=V\,\,units\]
\[{{V}_{2}}=?\]
\[P=m\times V\]
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