Answer
The magnitude of the impulse on particle 1 is $~~0.80~kg~m/s$
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 Equation (9-67) to find $v_{1f}$:
$v_{1f} = \frac{m_1-m_2}{m_1+m_2}~v_{1i}$
$v_{1f} = \frac{m_1-2~m_1}{m_1+2~m_1}~v_{1i}$
$v_{1f} = \frac{-m_1}{3~m_1}~v_{1i}$
$v_{1f} = -(\frac{1}{3})~(3.00~m/s)$
$v_{1f} = -1.00~m/s$
We can find the change in momentum of particle 1:
$\Delta p = m~\Delta v$
$\Delta p = m~(v_f-v_i)$
$\Delta p = (0.200~kg)~(-1.00~m/s-3.00~m/s)$
$\Delta p = -0.80~kg~m/s$
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 $~~0.80~kg~m/s$
