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

Two particles of masses m_1 and m_2 in projectile motion have velociti

1.	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
Answer» Zero `(m_1 +m_2)g t_(0)` `2(m_1 +m_2)g t_0` `(1)/(2)(m_1 +m_2)g t_0` Solution :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`

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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

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