A fire work rocket is fired vertically upward. At its maximum height of 100.0 m, it explodes and breaks into two pieces, one with mass 1.40 kg and...

Sunday, August 11, 2013

A fire work rocket is fired vertically upward. At its maximum height of 100.0 m, it explodes and breaks into two pieces, one with mass 1.40 kg and...

Denote the (positive) speeds of objects just after explosion as $\displaystyle{V}_{{1}}$ and $\displaystyle{V}_{{2}}$ , and their masses as $\displaystyle{m}_{{1}}$ and $\displaystyle{m}_{{2}}$ . The directions of $\displaystyle{V}_{{1}}$ and $\displaystyle{V}_{{2}}$ are the opposite.

Then conservation of impulse gives us $\displaystyle{m}_{{1}}{V}_{{1}}={m}_{{2}}{V}_{{2}},$ because the speed just before the explosion was zero. The conservation of energy gives $\displaystyle{m}_{{1}}{{V}_{{1}}^{{2}}}+{m}_{{2}}{{V}_{{2}}^{{2}}}={J}$ (the chemical energy that becomes kinetic).

Solving this system of equations is easy:

$\displaystyle{V}_{{1}}=\frac{{m}_{{2}}}{{m}_{{1}}}{V}_{{2}},$ substitute it to...

Solving this system of equations is easy:

$\displaystyle{V}_{{1}}=\frac{{m}_{{2}}}{{m}_{{1}}}{V}_{{2}},$ substitute it to the second equation and obtain $\displaystyle\frac{{m}_{{2}}}{{m}_{{1}}}{\left({m}_{{2}}+{m}_{{1}}\right)}{{V}_{{2}}^{{2}}}={2}{J},$ so

$\displaystyle{{V}_{{2}}^{{2}}}={2}{J}\frac{{m}_{{1}}}{{m}_{{2}}}\frac{{1}}{{{m}_{{1}}+{m}_{{2}}}}$ and $\displaystyle{{V}_{{1}}^{{2}}}={2}{J}\frac{{m}_{{2}}}{{m}_{{1}}}\frac{{1}}{{{m}_{{1}}+{m}_{{2}}}}.$

In numbers, they will be equal to

$\displaystyle{V}_{{1}}=\sqrt{{\frac{{1800}}{{1.7}}\frac{{0.3}}{{1.4}}}}\approx{15}{\left(\frac{{m}}{{s}}\right)},$ $\displaystyle{V}_{{2}}=\sqrt{{\frac{{1800}}{{1.7}}\frac{{1.4}}{{0.3}}}}\approx{70}{\left(\frac{{m}}{{s}}\right)}.$

For the second part of the question we have to consider an angle between the velocity of the first object and a horizontal line, and write the equations of horizontal and vertical movements of both bodies.

at August 11, 2013

Notes

Sunday, August 11, 2013