Answer
(a) After the collision, car A has a velocity of 3.65 m/s
After the collision, car B has a velocity of 4.45 m/s
(b) $\Delta p_A = -370~kg\cdot m/s$
$\Delta p_B = 370~kg\cdot m/s$
Work Step by Step
(a) Let $m_A = 435~kg$ and let $m_B = 495~kg$.
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 7-7 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:
$m_A~v_A + m_B~v_B = m_A~v_B' - m_Av_A + m_Av_B+ m_B~v_B'$
$v_B' = \frac{2m_A~v_A+m_B~v_B -m_Av_B}{m_A+m_B}$
$v_B’ = \frac{(2)(435~kg)(4.50~m/s)+(495~kg)(3.70~m/s)- (435~kg)(3.70~m/s)}{(435~kg)+(495~kg)}$
$v_B' = 4.45~m/s$
We can use $v_B'$ to find $v_A'$:
$v_A' = v_B' - v_A + v_B$
$v_A' = 4.45~m/s - 4.50~m/s + 3.70~m/s = 3.65~m/s$
(b)$\Delta p_A = m(v_A'-v_A)$
$\Delta p_A = (435~kg)(3.65~m/s - 4.50~m/s)$
$\Delta p_A = -370~kg\cdot m/s$
$\Delta p_B = m(v_B'-v_B)$
$\Delta p_B = (495~kg)(4.45~m/s - 3.70~m/s)$
$\Delta p_B = 370~kg\cdot m/s$
