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 `V_1` and `V_2` , and their masses as `m_1` and `m_2` . The directions of `V_1` and `V_2` are the opposite.


Then conservation of impulse gives us `m_1 V_1=m_2 V_2,` because the speed just before the explosion was zero. The conservation of energy gives `m_1 V_1^2+m_2 V_2^2=J` (the chemical energy that becomes kinetic).


Solving this system of equations is easy:


`V_1=m_2/m_1 V_2,` substitute it to...

Denote the (positive) speeds of objects just after explosion as `V_1` and `V_2` , and their masses as `m_1` and `m_2` . The directions of `V_1` and `V_2` are the opposite.


Then conservation of impulse gives us `m_1 V_1=m_2 V_2,` because the speed just before the explosion was zero. The conservation of energy gives `m_1 V_1^2+m_2 V_2^2=J` (the chemical energy that becomes kinetic).


Solving this system of equations is easy:


`V_1=m_2/m_1 V_2,` substitute it to the second equation and obtain `m_2/m_1 (m_2+m_1) V_2^2=2 J,` so


`V_2^2=2 J m_1/m_2 1/(m_1+m_2)` and `V_1^2=2 J m_2/m_1 1/(m_1+m_2).`


In numbers, they will be equal to


`V_1=sqrt(1800/1.7 0.3/1.4) approx 15(m/s),` `V_2=sqrt(1800/1.7 1.4/0.3) approx 70(m/s).`


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.


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