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 air. The value of $|({m_1}\overrightarrow {{v_1}} '\, + {m_2}\overrightarrow {{v_2}} ') - ({m_1}\overrightarrow {{v_1}} \, + {m_2}\overrightarrow {{v_2}} )$| is
Zero
$({m_1} + {m_2})g{t_0}$
$2({m_1} + {m_2})g{t_0}$
$\frac{1}{2}({m_1} + {m_2})g{t_0}$
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 air. The value of $|({m_1}\overrightarrow {{v_1}} '\, + {m_2}\overrightarrow {{v_2}} ') - ({m_1}\overrightarrow {{v_1}} \, + {m_2}\overrightarrow {{v_2}} )$| is
The momentum of the two-particle system, at $t = 0$ is
${\vec P_i} = {m_1}{\vec v_1} + {m_2}{\vec v_2}$
Collision between the two does not affect the total momentum of the system.
A constant external force $({m_1} + {m_2})g$ acts on the system.
The impulse given by this force, in time $t = 0$ to $t = 2{t_0}$ is $({m_1} + {m_2})g \times 2{t_0}$
|Change in momentum in this interval
$ = \,|{m_1}\vec v{'_1} + {m_2}\vec v{'_2} - ({m_1}{\vec v_1} + {m_2}{\vec v_2})|\, = 2({m_1} + {m_2})g{t_0}$
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