Problem 95 - University Physics with Modern Physics

Question

In a shipping company distribution center, an open cart of mass 50.0 \(\mathrm{kg}\) is rolling to the left at a speed of 5.00 \(\mathrm{m} / \mathrm{s}\) (Fig. P8.95). You can ignore friction between the cart and the floor. A 15.0 -kg package slides down a chute that is inclined at \(37^{\circ}\) from the horizontal and leaves the end of the chute with a speed of 3.00 \(\mathrm{m} / \mathrm{s}\) . The package lands in the cart and they roll off together. If the lower end roll off together. If the lower end of the chute is a vertical distance of 4.00 \(\mathrm{m}\) above the bottom of the cart, what are (a) the speed of the package just before it lands in the cart and (b) the final speed of the cart?

Step-by-Step Solution

Answer

The package's speed just before landing is \(9.36\ \mathrm{m/s}\) and the cart's final speed is \(-3.29\ \mathrm{m/s}\), moving to the left.

Step 1: Determine the vertical velocity of the package

Use the equation for vertical motion to find the velocity component of the package. Since the final vertical position is the height from which it falls, apply the formula: \[ v_{y}^2 = v_{0y}^2 + 2g h \]where:- \(v_{y}\) is the vertical component of final velocity,- \(v_{0y} = 3.00 \times \sin(37^{\circ})\) is the initial vertical velocity component,- \(g = 9.81 \ \mathrm{m/s^2}\) is the acceleration due to gravity,- \(h = 4.00 \ \mathrm{m}\) is the height.Calculate \(v_{0y}\) first:\[ v_{0y} = 3.00 \times \sin(37^{\circ}) \approx 1.80 \ \mathrm{m/s} \]Now calculate the vertical velocity:\[ v_{y}^2 = (1.80)^2 + 2 \times 9.81 \times 4.00 \]Solving gives:\[ v_{y} = \sqrt{1.80^2 + 78.48} = \sqrt{81.7} \approx 9.04 \ \mathrm{m/s} \]

Step 2: Calculate the horizontal component of the package's velocity

Find the horizontal velocity component of the package as it leaves the chute. This is determined using:\[ v_{0x} = 3.00 \times \cos(37^{\circ}) \]Calculate \(v_{0x}\):\[ v_{0x} = 3.00 \times \cos(37^{\circ}) \approx 2.40 \ \mathrm{m/s} \]

Step 3: Calculate the speed of the package just before landing in the cart

Combine the horizontal and vertical velocity components using the Pythagorean Theorem:\[ v_{total} = \sqrt{v_{0x}^2 + v_{y}^2} \]Calculate using the components:\[ v_{total} = \sqrt{(2.40)^2 + (9.04)^2} = \sqrt{5.76 + 81.7} \approx 9.36 \ \mathrm{m/s} \]

Step 4: Apply Conservation of Momentum to find final speed of the cart

Use the principle of conservation of momentum since no external forces act horizontally:\[ m_c v_{c} + m_p v_{px} = (m_c + m_p)v_f \]Where:- \(m_c = 50.0 \ \mathrm{kg}\) is the mass of the cart,- \(v_c = -5.00 \ \mathrm{m/s}\) is the velocity of the cart (negative as it moves left),- \(m_p = 15.0 \ \mathrm{kg}\) is the mass of the package,- \(v_{px} = 2.40 \ \mathrm{m/s}\) is the horizontal velocity component of the package,- \(v_f\) is the final velocity of the system.Substitute known values:\[ 50.0(-5.00) + 15.0(2.40) = (50.0 + 15.0)v_f \]Solving for \(v_f\):\[ -250.0 + 36.0 = 65.0v_f \]\[ v_f = \frac{-214.0}{65.0} \approx -3.29 \ \mathrm{m/s} \]

Step 5: Conclusion

Thus, the speed of the package just before landing in the cart is approximately \(9.36\ \mathrm{m/s}\) and the final speed of the cart after the collision is \(-3.29\ \mathrm{m/s}\). The negative sign indicates the direction of motion remains to the left.

Key Concepts

Velocity Components

Understanding velocity components is crucial when analyzing motion, especially in two dimensions. When an object moves along an inclined plane, its overall velocity is split into two distinct components: a horizontal and a vertical component.

• The horizontal velocity component is determined by projecting the object's velocity onto the horizontal axis. Typically, we use a trigonometric function, like cosine, when the angle of the incline and the initial velocity are known: \( v_{0x} = v_0 \cdot \cos(\theta) \), where \( v_0 \) is the initial speed and \( \theta \) is the angle.
• The vertical velocity component involves the sine of the angle when resolving the velocity into components: \( v_{0y} = v_0 \cdot \sin(\theta) \). This component changes over time because of the influence of gravity.

Without these components, it would be difficult to accurately determine how an object behaves as it moves along an inclined surface.
Understanding that these components are independent, yet together define the motion, is key to solving related physics problems.

Projectile Motion

Projectile motion involves objects being thrown or propelled into the air at an angle. This motion is characterized by a parabolic trajectory, achieved due to the combined horizontal and vertical motions under the influence of gravity.

In the case of our package example, it was released at an angle down a chute, following projectile motion principles:

• Initially at the top of the chute, the package has some velocity components, \( v_{0x} \) and \( v_{0y} \), influencing its path of motion.
• Gravity acts on the vertical component, pulling the package downwards, which influences the speed as it descends.
• Overall motion can be solved using known kinematic equations that allow you to calculate the position, speed, and velocity of a projectile at any point along its path.

This knowledge simplifies complex motion into more manageable calculations, as seen when determining the final speed before landing in the cart, combining vertical and horizontal motions.

Inclined Plane Dynamics

Inclined planes are common systems in physics that help illustrate force, motion, and energy concepts. When an object slides down an inclined plane, several forces come into play.

• The plane's angle impacts how quickly an object accelerates, with a steeper incline usually resulting in greater acceleration.
• Gravity plays a significant role, where the component of gravitational force acting along the plane causes motion.

When the package leaves the chute, it is influenced by these inclined plane dynamics, adjusting its path before reaching the cart. Neglecting friction simplifies this analysis, focusing mainly on the components of forces along and perpendicular to the plane. By stripping down this force analysis, understanding the eventual interaction with the cart becomes much clearer. By optimizing these principles, finding solutions for complex motion problems becomes significantly easier.