Two particles of masses `m_1 and m_2` in projectile motion have velocities `vecv_1 and vecv_2` respectively at time `t=0.` They collide at time `t_0.` Their velocities become `vecv_1' and vecv_2'` at time `2t_0` while still moving in air. The value of `|(m_1vecv_1' + m_2vecv_2') - (m_1vecv_1 + m_2vecv_2)|` is

Options

A. Zero
B. `(m_1 +m_2)g t_(0)`
C. `2(m_1 +m_2)g t_0`
D. `(1)/(2)(m_1 +m_2)g t_0`

Answer

c

Explanation

Momentum of the system at `t=0 vec(p_1)=m_1vec(v_1)+m_2vec(v_2)`
At time `t=2t_0 vec(p_2)=m_1v_1+m_2v_2`
Rate of change of momentum = Force acting on the system = `(vec(p_2)-vec(p_1))/(2t_0)=(m_1+m_2)g`
Putting the values of `vecp_1` and `vecp_2`
`((m_1v_1+m_2v_2)-(m_1v_1+m_2v_2))/(2t_0)=(m_1+m_2)g`
`:. (m_1v_1+m_2+v_2_-(m_1v_1+m_2v_2)=2(m_1+m_2)"gt"_0`