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Chapter 8: Problem 87

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 \((\textbf{Fig. P8.87})\). 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 m/s. The package lands in the cart and they roll together. If the lower end of the chute is a vertical distance of 4.00 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?

Short Answer

Expert verified
(a) Speed of package before landing is 9.35 m/s. (b) Final speed of the cart is -3.15 m/s.

Step by step solution

01

Analyze the problem

We need to find two things: the speed of the package just before it hits the cart, and the final speed of the cart once the package has landed in it. First, we need to determine the vertical speed the package gains by falling 4.00 m down (gravity's effect).
02

Determine vertical speed due to gravity

Use the kinematic equation to find the vertical speed gained by the package: \[ v^2 = u^2 + 2gh \]where \( v \) is the final vertical velocity, \( u = 0 \) (initially no vertical velocity when it leaves the chute), \( g = 9.8 \, m/s^2 \) is the acceleration due to gravity, and \( h = 4.00 \, m \) is the height. Substitute the values: \[ v^2 = 0 + 2 \times 9.8 \times 4 = 78.4 \]Hence, \( v = \sqrt{78.4} = 8.85 \, m/s \) approximately.
03

Calculate the speed of the package before landing

Now find the resultant speed of the package right before landing using Pythagoras theorem, considering its horizontal and vertical components. Horizontal speed \( v_{hx} = 3.00 \, m/s \), Vertical speed \( v_{vy} = 8.85 \, m/s \). Use:\[ v = \sqrt{v_{hx}^2 + v_{vy}^2} = \sqrt{3^2 + 8.85^2} \]\[ v = \sqrt{9 + 78.4225} = \sqrt{87.4225} = 9.35 \, m/s \]
04

Calculate the final speed of the cart

Using the conservation of momentum principle, calculate the final speed after the collision: \[ m_1u_1 + m_2u_2 = (m_1 + m_2)v_{final} \]where: - \( m_1 = 50 \, kg \) (mass of the cart)- \( u_1 = -5.00 \, m/s \) (initial speed of the cart, direction matters)- \( m_2 = 15 \) kg (mass of the package)- \( u_2 = 3.00 \, m/s \) (initial horizontal speed of the package)The net horizontal momentum: \[ 50 \times (-5) + 15 \times 3 = 65v_{final} \]\[ -250 + 45 = 65v_{final} \]\[ v_{final} = -205 / 65 \approx -3.15 \, m/s \]The negative sign indicates the cart moves to the left.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Kinematic Equations
Kinematic equations are fundamental in physics as they describe the motion of objects without considering the causes of motion, such as forces. These equations help us understand how different factors like distance, velocity, and acceleration interplay over time. They are vital when analyzing projectile motion or objects under gravity's influence.
One essential kinematic equation is:
  • \( v^2 = u^2 + 2gh \)
In this equation, \( v \) is the final velocity, \( u \) is the initial velocity, \( g \) is the acceleration due to gravity (approximately 9.8 m/s虏 on Earth), and \( h \) is the height from which the object falls.
By using kinematic equations, we can determine the final speed of a package that falls a certain distance and gains vertical velocity due to gravity. Understanding these equations allows us to calculate precise values and predict the motion of objects.
Projectile Motion
Projectile motion describes the motion of an object projected into space under the influence of gravity. A key aspect of projectile motion is that it involves two independent components: horizontal and vertical motions.
  • The horizontal component (constant velocity) does not influence the vertical motion.
  • The vertical component is influenced by gravity, with the object accelerating downwards.
In our example, a package slides down a chute and becomes a projectile once it leaves the chute. It simultaneously maintains a horizontal velocity (3.00 m/s) and gains vertical velocity due to gravity. To find the total speed of the package before landing, both components are analyzed separately, then combined using the Pythagorean theorem:
  • \( v = \sqrt{v_{hx}^2 + v_{vy}^2} \)
This allows for calculating the resultant speed of the package right before it lands.
Resultant Velocity Calculation
When an object moves in two dimensions, such as our package sliding off the chute, it can have velocities in both horizontal and vertical directions. To find out how fast the object is moving overall, we calculate the resultant velocity.
The package has distinct horizontal and vertical velocities before landing in the cart. Understanding this resultant velocity involves a coordinated view of both velocity components.
  • Horizontal Velocity, \( v_{hx} = 3.00 \, m/s \)
  • Vertical Velocity, \( v_{vy} = 8.85 \, m/s \)
Using these, the resultant velocity is calculated using the formula:
  • \( v = \sqrt{v_{hx}^2 + v_{vy}^2} = \sqrt{3^2 + 8.85^2} \)
Thus, the resultant speed tells us how fast the object is traveling right at the moment it lands, combining its movement effects in both directions. This velocity is crucial in understanding how objects interact during collision or integration, like when the package lands in the cart.

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Most popular questions from this chapter

A machine part consists of a thin, uniform 4.00-kg bar that is 1.50 m long, hinged perpendicular to a similar vertical bar of mass 3.00 kg and length 1.80 m. The longer bar has a small but dense 2.00-kg ball at one end (\(\textbf{Fig. E8.55}\)). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through 90\(^\circ\) to make the entire part horizontal?

The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0-ms contact time. The fastest-known served tennis ball was served by "Big Bill" Tilden in 1931, and its speed was measured to be 73 m/s. (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of 55 m/s, what force and what impulse did he exert on the ball, assuming only horizontal motion?

Two cars collide at an intersection. Car \(A\), with a mass of 2000 kg, is going from west to east, while car \(B\), of mass 1500 kg, is going from north to south at 15 m/s. As a result, the two cars become enmeshed and move as one. As an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of 65\(^\circ\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

A railroad handcar is moving along straight, frictionless tracks with negligible air resistance. In the following cases, the car initially has a total mass (car and contents) of \(200 \mathrm{~kg}\) and is traveling east with a velocity of magnitude \(5.00 \mathrm{~m} / \mathrm{s}\). Find the final velocity of the car in each case, assuming that the handcar does not leave the tracks. (a) A \(25.0-\mathrm{kg}\) mass is thrown sideways out of the car with a velocity of magnitude \(2.00 \mathrm{~m} / \mathrm{s}\) relative to the car's initial velocity. (b) A 25.0 -kg mass is thrown back ward out of the car with a velocity of \(5.00 \mathrm{~m} / \mathrm{s}\) relative to the initial motion of the car. (c) A 25.0 -kg mass is thrown into the car with a velocity of \(6.00 \mathrm{~m} / \mathrm{s}\) relative to the ground and opposite in direction to the initial velocity of the car.

A 5.00-g bullet is fired horizontally into a 1.20-kg wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is 0.20. The bullet remains embedded in the block, which is observed to slide 0.310 m along the surface before stopping. What was the initial speed of the bullet?
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