Answer
(a) The magnitude of the 10.0-g marble is 0.500 m/s, and it is moving to the right.
The magnitude of the 30.0-g marble is 0.100 m/s and, it is moving to the left.
(b) The change in momentum is equal in magnitude but in the opposite directions for the two marbles. The change in momentum for the 10.0-g marble is $0.00900~kg~m/s$, while the change in momentum for the 30.0-g marble is $-0.00900~kg~m/s$,
(c) The change in kinetic energy is equal in magnitude for the two marbles. However the 10.0-g marble gained 0.000450 J of kinetic energy, while the 30.0-g marble lost 0.000450 J of kinetic energy.
Work Step by Step
Let $v_A'$ be the final velocity of the 10.0-g marble.
Let $v_B'$ be the final velocity of the 30.0-g marble.
Let $m_A = 0.0100~kg$ and let $m_B = 0.0300~kg$
We can use conservation of momentum to set up an equation.
$m_A~v_A + m_B~v_B = m_A~v_A' + m_B~v_B'$
Since the collision is elastic, we can use equation (8.27) to set up another equation.
$v_A - v_B = v_B' - v_A'$
$v_A' = v_B' - v_A + v_B$
We can use this expression for $v_A'$ in the first equation.
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B’ = \frac{(2)(0.0100~kg)(-0.400~m/s)+(0.0300~kg)(0.200~m/s)- (0.0100~kg)(0.200~m/s)}{(0.0100~kg)+(0.0300~kg)}$
$v_B' = -0.100~m/s$
We can use $v_B'$ to find $v_A'$.
$v_A' = v_B' - v_A + v_B$
$v_A' = -0.100~m/s - (-0.400)~m/s + 0.200~m/s$
$v_A' = 0.500~m/s$
The magnitude of the 10.0-g marble is 0.500 m/s and it is moving to the right.
The magnitude of the 30.0-g marble is 0.100 m/s and it is moving to the left.
(b) $\Delta p_A = (0.0100~kg)(0.500~m/s)-(0.0100~kg)(-0.400~m/s)$
$\Delta p_A = 0.00900~kg~m/s$
$\Delta p_B = (0.0300~kg)(-0.100~m/s)-(0.0300~kg)(0.200~m/s)$
$\Delta p_B = -0.00900~kg~m/s$
The change in momentum is equal in magnitude but in the opposite directions for the two marbles. The change in momentum for the 10.0-g marble is $0.00900~kg~m/s$ while the change in momentum for the 30.0-g marble is $-0.00900~kg~m/s$
(c) $\Delta K_A = K_2-K_1$
$\Delta K_A = \frac{1}{2}(0.0100~kg)(0.500~m/s)^2-\frac{1}{2}(0.0100~kg)(0.400~m/s)^2$
$\Delta K_A = 0.000450~J$
$\Delta K_B = K_2-K_1$
$\Delta K_B = \frac{1}{2}(0.0300~kg)(0.100~m/s)^2-\frac{1}{2}(0.0300~kg)(0.200~m/s)^2$
$\Delta K_B = -0.000450~J$
The change in kinetic energy is equal in magnitude for the two marbles. However the 10.0-g marble gained 0.000450 J of kinetic energy while the 30.0-g marble lost 0.000450 J of kinetic energy.
