The force on a rocket moving with a velocity $$300\,m/s$$ is $$210\,N.$$ The rate of consumption of fuel of rocket is
A.
$$0.7\,kg/s$$
B.
$$1.4\,kg/s$$
C.
$$0.07\,kg/s$$
D.
$$10.7\,kg/s$$
Answer :
$$0.7\,kg/s$$
Solution :
Whenever there is change in the mass w.r.t. time, apply $$F = - v\frac{{dm}}{{dt}}$$
Thrust force on the rocket $${F_t} = {v_r}\left( { - \frac{{dm}}{{dt}}} \right)\,\,\left( {{\text{upwards}}} \right)$$
Rate of combustion of fuel $$ - \frac{{dm}}{{dt}} = \frac{{{F_t}}}{{{v_r}}}$$
Given, $${F_t} = 210\,N$$
$$\eqalign{
& {v_r} = 300\,m/s \cr
& \therefore - \frac{{dm}}{{dt}} = \frac{{210}}{{300}} = 0.7\,kg/s \cr} $$
Releted MCQ Question on Basic Physics >> Momentum
Releted Question 1
Two particles of masses $${m_1}$$ and $${m_2}$$ in projectile motion have velocities $${{\vec v}_1}$$ and $${{\vec v}_2}$$ respectively at time $$t = 0.$$ They collide at time $${t_0.}$$ Their velocities become $${{\vec v}_1}'$$ and $${{\vec v}_2}'$$ at time $$2{t_0}$$ while still moving in the air. The value of $$\left| {\left( {{m_1}{{\vec v}_1}' + {m_2}{{\vec v}_2}'} \right) - \left( {{m_1}{{\vec v}_1} + {m_2}{{\vec v}_2}} \right)} \right|$$ is
A.
zero
B.
$$\left( {{m_1} + {m_2}} \right)g{t_0}$$
C.
$$\frac{1}{2}\left( {{m_1} + {m_2}} \right)g{t_0}$$
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