Answer
(a) The magnitude of the velocity is 6.44 m/s and they are moving to the east after the collision.
(b) The truck should have been moving with a speed of 2.49 m/s.
(c) The change in kinetic energy is greater in magnitude in part (a).
Work Step by Step
(a) We can use conservation of momentum to solve this question. Let the east be the positive direction.
$(m_A+m_B)~v_2 = m_A~v_{A1}+m_B~v_{B1}$
$v_2 = \frac{m_A~v_{A1}+m_B~v_{B1}}{m_A+m_B}$
$v_2 = \frac{(1050~kg)(-15.0~m/s)+(6320~kg)(10.0~m/s)}{1050~kg+6320~kg}$
$v_2 = 6.44~m/s$
The magnitude of the velocity is 6.44 m/s, and they are moving to the east after the collision.
(b) In this case, the total momentum in the system is zero. Then the momentum of the two vehicles would be equal in magnitude.
$m_B~v_{B1} = m_A~v_{A1}$
$v_{B1} = \frac{m_A~v_{A1}}{m_B}$
$v_{B1} = \frac{(1050~kg)(15.0~m/s)}{6320~kg}$
$v_{B1} = 2.49~m/s$
The truck should have been moving with a speed of 2.49 m/s.
(c) We can find $K_1$ in part (a):
$K_1 = \frac{1}{2}m_A~v_A^2+\frac{1}{2}m_B~v_B^2$
$K_1 = \frac{1}{2}(1050~kg)(15.0~m/s)^2+\frac{1}{2}(6320~kg)(10.0~m/s)^2$
$K_1 = 434125~J$
We can find $K_2$ for part (a)
$K_2 = \frac{1}{2}m_2~v_2^2$
$K_2 = \frac{1}{2}(7370~kg)(6.44~m/s)^2$
$K_2 = 152830~J$
$\Delta K = K_2 - K_1$
$\Delta K = 152830~J - 434125~J$
$\Delta K = -281295~J$
We can find $K_1$ in part (b)
$K_1 = \frac{1}{2}m_A~v_A^2+\frac{1}{2}m_B~v_B^2$
$K_1 = \frac{1}{2}(1050~kg)(15.0~m/s)^2+\frac{1}{2}(6320~kg)(2.49~m/s)^2$
$K_1 = 137717~J$
In part (b), $K_2 = 0$
$\Delta K = K_2 - K_1$
$\Delta K = 0 - 137717~J$
$\Delta K = -137717~J$
We can see that the change in kinetic energy is greater in magnitude in part (a).
