/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 111 A rocket sled with a mass of \(2... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 111

A rocket sled with a mass of \(2900 \mathrm{~kg}\) moves at \(250 \mathrm{~m} / \mathrm{s}\) on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after \(920 \mathrm{~kg}\) of water has been scooped up. Ignore any retarding force on the scoop.

Short Answer

Expert verified
The speed of the sled after scooping water is approximately 189.79 m/s.

Step by step solution

01

Understand the conservation of momentum

The principle of conservation of linear momentum states that if no external forces are acting on a system, the total momentum of the system remains constant. In this case, the rocket sled and the water it scoops up form a closed system.
02

Write down the initial momentum

Initial momentum is the product of the initial mass and initial velocity of the sled. The formula is given by: \( p_{initial} = m_{sled} \cdot v_{initial} \). Given \( m_{sled} = 2900 \, \text{kg} \) and \( v_{initial} = 250 \, \text{m/s} \), we have:\[ p_{initial} = 2900 \, \text{kg} \times 250 \, \text{m/s} = 725,000 \, \text{kg} \cdot \text{m/s} \]
03

Determine the final mass of the sled and water

After the sled scoops up the water, the total mass increases by the mass of the water. Thus, the final mass \( m_{final} \) is:\[ m_{final} = m_{sled} + m_{water} = 2900 \, \text{kg} + 920 \, \text{kg} = 3820 \, \text{kg} \]
04

Write down the final momentum

The final momentum of the system (sled and water) is given by the product of the final mass and the final velocity of the sled (which we need to find). So, \( p_{final} = m_{final} \cdot v_{final} \).
05

Equate initial and final momentum to solve for final velocity

By the conservation of momentum, \( p_{initial} = p_{final} \). Therefore:\[ 725,000 \, \text{kg} \cdot \text{m/s} = 3820 \, \text{kg} \cdot v_{final} \]Now solve for \( v_{final} \):\[ v_{final} = \frac{725,000 \, \text{kg} \cdot \text{m/s}}{3820 \, \text{kg}} \approx 189.79 \, \text{m/s} \]
06

Conclusion

After the sled scoops up the water, its speed is approximately \(189.79 \, \text{m/s}\).

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Key Concepts

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

Linear Momentum
Linear momentum is a crucial concept in physics that helps us understand how objects move and interact in space. It is defined as the product of an object's mass and its velocity. Mathematically, this is expressed as:\[ p = m \times v \]where:
  • \(p\) is the linear momentum,
  • \(m\) is the mass of the object, and
  • \(v\) is the object's velocity.
Linear momentum is a vector quantity. This means it has both magnitude and direction, aligning with the direction of the object's velocity.
If no external forces act on an object or system of objects, the total linear momentum is conserved. This is the principle of conservation of momentum.
In the rocket sled problem, the conservation of linear momentum allows us to predict the sled's speed after scooping up water without considering external forces, making it a powerful tool for solving such problems.
Rocket Sled
A rocket sled is a testing device that uses rockets to propel itself along a set of rails. It's essentially like a train, but much faster and used primarily for experimental purposes.
The rocket sled in our exercise is moving at a high speed on rails while scooping up additional mass (water) from the ground as part of the experimental setup. This setup demonstrates the application of the conservation of momentum by changing the mass of the sled mid-motion. As it scoops up water, the sled's momentum adjusts according to the principles of physics, allowing us to calculate new velocities easily.
Rocket sleds are crucial in testing aerodynamics, material strength, and other engineering questions, particularly in military and aerospace applications. They help perform tests safely and accurately without the need to launch actual rockets into the sky each time.
Mass-Velocity Relationship
Understanding the mass-velocity relationship is key to solving problems involving momentum. In the context of the rocket sled, this relationship shows how mass and velocity interplay to affect the sled's movement.
When mass increases鈥攕uch as when the sled scoops water鈥攖he sled's velocity must decrease if there are no external forces; all due to the conservation of linear momentum. Let's break it down:
  • Initially, the sled has a certain mass (2900 kg) and velocity (250 m/s).
  • When it scoops up 920 kg of water, its mass becomes 3820 kg.
The initial momentum is calculated with its first mass and velocity.
We use momentum conservation to find the new velocity once we know the new mass.
So, the larger the mass becomes, the smaller the velocity becomes if momentum stays the same. This is why the sled slows down after collecting the water.

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Most popular questions from this chapter

A steel ball of mass \(0.500 \mathrm{~kg}\) is fastened to a cord that is \(70.0 \mathrm{~cm}\) long and fixed at the far end. The ball is then released when the cord is horizontal (Fig. \(9-65\) ). At the bottom of its path, the ball strikes a \(2.50 \mathrm{~kg}\) steel block initially at rest on a frictionless surface. The collision is elastic. Find (a) the speed of the ball and (b) the speed of the block, both just after the collision.

puck 1 of mass \(m_{1}=0.20 \mathrm{~kg}\) is sent sliding across a frictionless lab bench, to undergo a one-dimensional elastic collision with stationary puck \(2 .\) Puck 2 then slides off the bench and lands a distance \(d\) from the base of the bench. Puck 1 rebounds from the collision and slides off the opposite edge of the bench. landing a distance \(2 d\) from the base of the bench. What is the mass of puck \(2 ?\) (Hint: Be careful with signs.)

Shows a two-ended "rocket" that is initially stationary on a frictionless floor, with its center at the origin of an \(x\) axis. The rocket consists of a central block \(C\) (of mass \(M=6.00 \mathrm{~kg}\) ) and blocks \(L\) and \(R\) (cach of mass \(m=2.00 \mathrm{~kg}\) ) on the left and right sides. Small explosions can shoot either of the side blocks away from block \(C\) and along the \(x\) axis. Here is the sequence: (1) At time \(t=0,\) block \(L\) is shot to the left with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that the explosion gives the rest of the rocket. (2) Next, at time \(t=0.80 \mathrm{~s},\) block \(R\) is shot to the right with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\) relative to the velocity that block \(C\) then has. At \(t=2.80 \mathrm{~s},\) what are (a) the velocity of block \(C\) and (b) the position of its center?

Particle \(A\) and particle \(B\) are held together with a com- pressed spring between them. When they are released, the spring pushes them apart, and they then fly off in opposite directions, free of the spring. The mass of \(A\) is 2.00 times the mass of \(B,\) and the energy stored in the spring was 60 J. Assume that the spring has negligible mass and that all its stored energy is transferred to the particles. Once that transfer is complete, what are the kinetic energics of (a) particle \(A\) and (b) particle \(B ?\)

In a common but dangcrous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\). and the collision on the floor lasts \(0.082 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?
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