Question

In a shipping company distribution center, an open cart of mass 50.0 kg is rolling to the left at a speed of 5.00 m/s. You can ignore friction between the cart and the floor. A 15.0 kg package slides down a chute that is inclined at 37° from the horizontal and leaves the end of the chute with a speed of 3.00 m/s. The package lands in the cart and they roll off together. If the lower end of the chute is a vertical distance of 4.00 m above the bottom of the cart, what are the speed of the package just before it lands in the cart and what is the final speed of the cart?

Answer 1

Explanations & Calculations

• To calculate the final velocity of the cart-package system, consider the conservation of linear momentum of the system leftward.

$\qquad\qquad \begin{aligned} \leftarrow\\ \small m_1v_1+m_2v_2&=\small (m_1+m_2)V\\ \small V&=\small \frac{50kg\times5ms^{-1}+15kg\times3ms^{-1}\cos37}{(50+15)kg} \\ &=\small 4.4\,ms^{-1} \end{aligned}$

• The package experiences a projectile just after leaving the chute. Therefore, it has both vertical and horizontal velocities. The horizontal velocity is already been used in the previous part which remains the same for the entire projectile.
• So to calculate the vertical velocity component just before the package lands in the cart, apply $\small v^2=u^2+2as$.

$\qquad\qquad \begin{aligned} \small v^2&=\small (3\sin37)^2+2\times9.8\times4\\ \small v&=\small 9.0ms^{-1} \end{aligned}$

• Therefore, the velocity just before landing in the cart is

$\qquad\qquad \begin{aligned} \small v_f&=\small \sqrt{9^2+(3\cos37))^2}\\ &=\small 9.3ms^{-1} \end{aligned}$

• Its direction is

$\qquad\qquad \begin{aligned} \small \theta&=\small \tan^{-1}\Big(\frac{9}{3\cos37}\Big)\\ &=\small 75.1^0[South\,of\,West] \end{aligned}$