At the top of its flight, it explodes, breaking into two parts, one of which has twice the mass of the other. The two particles land simultaneously. The lighter fragment lands back at launch point. Where does the other fragment land? What is the energy of explosion?
Velocity at top of flight $$=35\cos 60$$ $$=\frac{35}{2}m/s$$
Using momentum conservation $$(6)(\frac{35}{2})=(2)(\frac{-35}{2})+4v$$ $$v=35$$
Also distance from the origin at point of explosion $$d=\frac{u^2\sin2\theta}{2g}$$ $$d=\frac{245\sqrt 3}{4}$$ Distance covered by 4kg mass from the point of explosion in time $$T=\frac{2u\sin\theta}{2g}$$ $$=\frac{7\sqrt 3}{4}$$ distance covered $$x=\frac{245\sqrt 3}{4}$$
So total distance is $$d+x$$ $$\frac{245\sqrt 3}{2}$$ But the answer given is is 160m. What am I doing wrong?
There's an error in the step where you obtain $d$ $$d = \frac{u^2 \sin{2\theta}}{2g} = \frac {1225 \times \sqrt{3}}{2\times 2 \times 10} = \frac{245 \sqrt{3}}{8}$$
on calculating $d+x$, we obtain $$d+x = \frac {245 \times 3 \sqrt 3}{8} \approx 160$$ which is the answer.