A machine gun fires a bullet of mass $$40g$$ with a velocity $$1200m{s^{ - 1}}.$$ The man holding it can exert a maximum force of $$144N$$ on the gun. How many bullets can he fire per second at the most?
A.
Two
B.
Four
C.
One
D.
Three
Answer :
Three
Solution :
Let $$n$$ be the number of bullets that the man can fire in one second.
$$\therefore $$ change in momentum per second $$ = n \times mv = F$$
[ $$m$$ = mass of bullet, $$v$$ = velocity] ($$\because F$$ is the force)
$$\therefore n = \frac{F}{{mv}} = \frac{{144 \times 1000}}{{40 \times 1200}} = 3$$
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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