Answer:
Here, \[m=5kg\], \[\upsilon
=20m{{s}^{-1}}\], \[\theta ={{60}^{o}}\]
Horizontal compt. Of velocity, \[{{\upsilon }_{x}}=\upsilon \cos
{{60}^{o}}=20\times \frac{1}{2}=10m/s\]
Vertical compt. Of velocity, \[{{\upsilon }_{y}}=\upsilon \sin
{{60}^{o}}=20\times \frac{\sqrt{3}}{2}=10\sqrt{3}m/s\]
Time taken to reach the highest point = time taken to reach the
ground from the highest point
\[t=\frac{\upsilon \sin \theta }{g}=\frac{{{\upsilon
}_{v}}}{g}=\frac{10\sqrt{3}}{9.8}=1.77s\]
At the highest point, m splits up into two parts of masses \[{{m}_{1}}=1kg\]and\[{{m}_{2}}=4kg\].
If their velocities \[{{\upsilon }_{1}}\]are\[{{\upsilon }_{2}}\] respectively,
then applying the principle of conservation of linear momentum, we get
\[{{m}_{1}}{{\upsilon }_{1}}+{{m}_{2}}{{\upsilon }_{2}}=m\upsilon
\cos \theta \]
\[{{\upsilon }_{1}}+4{{\upsilon }_{2}}=5\times 10=50\] ??.
(i)
Initial K.E. = \[\frac{1}{2}m{{\left( \upsilon \cos \theta
\right)}^{2}}=\frac{1}{2}\times 5{{\left( 10 \right)}^{2}}=250J\]
Final K.E. = 2 (initial K.E.)= \[2\times 250=500J\]
\[\therefore \]\[\frac{1}{2}{{m}_{1}}\upsilon
_{1}^{2}+\frac{1}{2}{{m}_{2}}\upsilon _{2}^{2}=500\]
\[\frac{1}{2}\times 1\upsilon
_{1}^{2}+\frac{1}{2}4\upsilon _{2}^{2}=500\] or \[\upsilon
_{1}^{2}+4\upsilon _{2}^{2}=1000\] ?? (ii)
On solving (i) and (ii), we get \[{{\upsilon }_{1}}=30m/s\],
\[{{\upsilon }_{2}}=5m/s\]
\[\therefore \] Separation between the two fragments = \[({{\upsilon
}_{1}}-{{\upsilon }_{2}})\times t=\left( 30-5 \right)\times
1.77metre\,=44.25\,m\]
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