question_answer

An object of mass 5kg is projected with a velocity of \[20\,m{{s}^{-1}}\]an angle \[{{60}^{o}}\]to the horizontal. At the highest point of its path, the projectile explodes and breaks up into two fragments of masses 1 kg and 4 kg. The fragments separate horizontally after the explosion, which releases internal energy such that K.E. of the system at the highest point is doubled. Calculate the separation between the two fragments when they reach the ground.

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 = \[({{\upsilon }_{1}}-{{\upsilon }_{2}})\times t=\left( 30-5 \right)\times 1.77metre\,=44.25\,m\]