An object of mass 5 kg is projected with a velocity 20ms^ - 1 at an...

Given, $m = 5kg, v = 20ms^{- 1}, \theta = 60 °$

Vertical component of velocity, ${v}_{y} = vsin60 °$

$= 20 \times \dfrac{3}{2} = 10\sqrt{3}ms^{- 1}$

Time taken to reach the highest point = Time taken to reach the ground from highest point.

$t = \dfrac{vsin\theta}{g} = \dfrac{vy}{g} = \dfrac{10\sqrt{3}}{9 . 8} = 1 . 77s$

At the highest point, $m$ splits up into two parts of masses ${m}_{1} = 1kg$ and ${m}_{2} = 4kg$

If their velocities are ${v}_{1}$ and ${v}_{2}$ respectively, then applying the principle of conservation of linear momentum, we get

${m}_{1}{v}_{1} + {m}_{2}{v}_{2} = mvcos\theta$

${v}_{1} + 4{v}_{2} = 5 \times 20 \times \dfrac{1}{2} \left[∵ \theta =60 °\right]$

${v}_{1} + 4{v}_{2} = 5 \times10 = 50 . . . . \left(i\right)$

Initial $KE =$\d$dfrac{1}{2}m\lleft(vcos\theta)2$

$= \dfrac{1}{2} \times5 \times \left(10\right)2 = 250J$

Final $KE = 2$ (initial $KE$)$= 2 \times250 = 500J$

$∴ \dfrac{1}{2}{m}_{1}{v}_{1}^{2} + \dfrac{1}{2}{m}_{2}{v}_{2}^{2} =500$

or $\dfrac{1}{2} \times1 \times {v}_{1}^{2} + \dfrac{1}{2} \times4 \times {v}_{2}^{2} =500$

or ${v}_{1}^{2} + 4{v}_{2}^{2} = 1000 . . . . \left(ii\right)$

Solving Eq. (i) and (ii), we get

${v}_{1} = 30ms^{- 1}, {v}_{2} = 5ms^{- 1}$

Hence, the separation between the two fragments

$= \left({v}_{1} - {v}_{2}\right) \timest = \left(30 - 5\right) \times1 . 77m = 44 . 25m</p>$