Answer
The magnitude of the impulse on particle 1 is $~~0.40~kg~m/s$
Work Step by Step
Let the initial velocity of particle 1 be $v_{1i} = 3.00~m/s$
Note that $~~m_2 = 2~m_1$
We can use conservation of momentum to find the velocity $v_f$ after the collision:
$p_f = p_i$
$(m_1+m_2)~v_f = m_1~(3.00~m/s)$
$(m_1+2~m_1)~v_f = m_1~(3.00~m/s)$
$(3~m_1)~v_f = m_1~(3.00~m/s)$
$3~v_f = 3.00~m/s$
$v_f = 1.00~m/s$
We can find the change in momentum of particle 1:
$\Delta p = m~\Delta v$
$\Delta p = m~(v_f-v_i)$
$\Delta p = (0.200~kg)~(1.00~m/s-3.00~m/s)$
$\Delta p = -0.40~kg~m/s$
The impulse on particle 1 is equal to the change in the momentum of particle 1.
Therefore, the magnitude of the impulse on particle 1 is $~~0.40~kg~m/s$
