Answer:
Refer to Fig. 4(HT). 10. Applying principle of
conservation of linear momentum to
(i) x component of motion
\[mV=2mV_{x}^{'}\] \[\therefore \] \[V_{x}^{'}=V/2\]
(ii) y component of motion
\[V'=\sqrt{V_{x}^{'2}+V_{y}^{'2}}=\sqrt{{{\left( V/2
\right)}^{2}}+{{\left( V/2 \right)}^{2}}}=\frac{V}{2}\sqrt{2}=V/\sqrt{2}\]
\[\tan \theta
=\frac{V_{y}^{'}}{V_{x}^{'}}=\frac{V/2}{V/2}=1\] \[\therefore
\]\[\theta ={{45}^{o}}\]
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