Answer
(a) The speed of the package just before it lands in the cart is 9.35 m/s.
(b) The final speed of the cart is 3.29 m/s.
Work Step by Step
(a) We can use conservation of energy to find the speed of the package after it falls 4.00 meters.
$K_2+U_2 = K_1+U_1$
$\frac{1}{2}mv_2^2+0 = \frac{1}{2}mv_1^2+mgh$
$v_2^2 = v_1^2+2gh$
$v_2 = \sqrt{v_1^2+2gh}$
$v_2 = \sqrt{(3.00~m/s)^2+(2)(9.80~m/s^2)(4.00~m)}$
$v_2 = 9.35~m/s$
The speed of the package just before it lands in the cart is 9.35 m/s.
(b) We can use conservation of momentum to find the final speed of the cart.
$(m_c+m_p)~v_2 = m_c~v_{c1}+m_p~v_{p1}~cos(\theta)$
$v_2 = \frac{m_c~v_{c1}+m_p~v_{p1}~cos(\theta)}{m_c+m_p}$
$v_2 = \frac{(50.0~kg)(5.00~m/s)+(15.0~kg)(-3.00~m/s)~cos(37^{\circ})}{50.0~kg+15.0~kg}$
$v_2 = 3.29~m/s$
The final speed of the cart is 3.29 m/s.
