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Chapter 8: Problem 29

You (mass 55 kg) are riding a frictionless skateboard (mass 5.0 kg) in a straight line at a speed of 4.5 m/s. A friend standing on a balcony above you drops a 2.5-kg sack of flour straight down into your arms. (a) What is your new speed while you hold the sack? (b) Since the sack was dropped vertically, how can it affect your \(horizontal\) motion? Explain. (c) Now you try to rid yourself of the extra weight by throwing the sack straight up. What will be your speed while the sack is in the air? Explain.

Short Answer

Expert verified
(a) The new speed is 4.32 m/s. (b) The sack does not affect horizontal motion because momentum is conserved separately in vertical and horizontal directions. (c) Your speed remains 4.32 m/s while the sack is in the air.

Step by step solution

01

Understand the Conservation of Momentum

Since the system is isolated and frictionless, we will use the conservation of momentum to determine the velocity after catching the sack. The formula for momentum is given by \( p = mv \), where \( m \) is mass and \( v \) is velocity. Initially, you and the skateboard are moving together, and the sack has no horizontal momentum because it is falling vertically.
02

Calculate Initial Momentum

Initial momentum (before catching the sack) is calculated by multiplying the total mass of you and the skateboard by your initial velocity. \[ p_{initial} = (55 \, \text{kg} + 5.0 \, \text{kg}) \times 4.5 \, \text{m/s} \] \[ p_{initial} = 60 \, \text{kg} \times 4.5 \, \text{m/s} = 270 \, \text{kg} \cdot \text{m/s} \]
03

Calculate Final Momentum

After catching the sack, you, the skateboard, and the sack now have a combined mass. The final momentum must equal the initial momentum because no external horizontal forces are acting. \[ p_{final} = (55 \, \text{kg} + 5.0 \, \text{kg} + 2.5 \, \text{kg}) \times v_{final} \]
04

Solve for Final Velocity

Set the final momentum equal to the initial momentum to solve for the new speed \( v_{final} \). \[ 270 \, \text{kg} \cdot \text{m/s} = 62.5 \, \text{kg} \times v_{final} \] \[ v_{final} = \frac{270 \, \text{kg} \cdot \text{m/s}}{62.5 \, \text{kg}} = 4.32 \, \text{m/s} \]
05

Understand the Vertical Motion Impact

The vertical drop of the sack does not directly affect the horizontal momentum because momentum is a vector quantity and is conserved separately in each direction. The sack was only imparted vertical momentum when dropped, but it joined your existing horizontal motion when caught.
06

Calculate Velocity after Throwing the Sack

When throwing the sack straight up, only vertical momentum is influenced. The horizontal momentum remains constant because there is still no external horizontal force. Therefore, your horizontal speed remains \( 4.32 \, \text{m/s} \) while the sack is in the air.

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics that measures the motion of an object. It is essentially the product of an object's mass and velocity, expressed as \( p = mv \). This means that momentum depends on both how much mass an object has and how fast it is moving.
  • For instance, a heavier object moving at the same speed as a lighter one will have more momentum.
  • Likewise, an object moving faster will have more momentum than the same object moving slower.
In the context of this exercise, when a sack of flour is dropped vertically onto you while on a skateboard, the principle of conservation of momentum plays a crucial role. The system (you, the skateboard, and the sack) should conserve its total momentum in the absence of external forces, particularly in the horizontal direction. This means that the total horizontal momentum before and after catching the sack remains the same.
Therefore, when the sack joins your motion, it does not add to the horizontal momentum but redistributes the existing momentum over a now larger mass (your mass plus the sack's). This helps us compute the new velocity after catching the sack.
Horizontal Motion
Horizontal motion refers to any motion happening parallel to the horizon. Understanding how forces affect horizontal motion helps us predict how objects move.
In physics, horizontal motion is especially crucial when dealing with multiple dimensions, as each direction can be analyzed separately. In this problem, since the sack is dropped vertically, it initially carries no horizontal momentum. However, once you catch it, it becomes part of the moving system.
  • Even though it was dropped, its presence does not change the total horizontal momentum of the system.
  • This is because horizontal and vertical motions have independent momentum."
Therefore, catching the sack doesn't change the horizontal speed unless an external force acts on the system. This concept, known as decoupled motion, is vital for understanding two- or three-dimensional problems in physics.
Velocity
Velocity is the measure of how fast an object moves in a specific direction. It鈥檚 different from speed, which only tells us how fast something is moving regardless of direction.
In the exercise, your initial velocity was 4.5 m/s. After catching the sack of flour, you and your skateboard momentarily move together with the sack at a new velocity because of the added mass.
  • The velocity can be updated using the principle of conservation of momentum, indicating no external horizontal forces are at play.
  • The new velocity, calculated using the initial and final masses, comes out to 4.32 m/s.
Interestingly, when you throw the sack straight up, your horizontal velocity does not change (stays at 4.32 m/s) since the momentum in the horizontal direction remains unaffected. This is because the act of throwing the sack impacts only the vertical axis. In physics, such considerations help us understand object trajectories and motion behavior in detail.

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Most popular questions from this chapter

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 fireworks rocket is fired vertically upward. At its maximum height of 80.0 m, it explodes and breaks into two pieces: one with mass 1.40 kg and the other with mass 0.28 kg. In the explosion, 860 J of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, what fraction of the initial mass of the rocket and fuel is \(not\) fuel? (b) What is this fraction if the final speed is to be 3000 m/s?

In July 2005, \(NASA\)'s "Deep Impact" mission crashed a 372-kg probe directly onto the surface of the comet Tempel 1, hitting the surface at 37,000 km/h. The original speed of the comet at that time was about 40,000 km/h, and its mass was estimated to be in the range (0.10 - 2.5) \(\times\)10\(^{14}\) kg. Use the smallest value of the estimated mass. (a) What change in the comet's velocity did this collision produce? Would this change be noticeable? (b) Suppose this comet were to hit the earth and fuse with it. By how much would it change our planet's velocity? Would this change be noticeable? (The mass of the earth is 5.97 \(\times\) 10\(^{24}\) kg.)

A 15.0-kg block is attached to a very light horizontal spring of force constant 500.0 N>m and is resting on a frictionless horizontal table (\(\textbf{Fig. E8.44}\)). Suddenly it is struck by a 3.00-kg stone traveling horizontally at 8.00 m/s to the right, whereupon the stone rebounds at 2.00 m/s horizontally to the left. Find the maximum distance that the block will compress the spring after the collision.
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