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Chapter 7: Problem 2

A model rocket is constructed with a motor that can provide a total impulse of \(29.0 \mathrm{N} \cdot \mathrm{s}\) The mass of the rocket is 0.175 kg. What is the speed that this rocket achieves when launched from rest? Neglect the effects of gravity and air resistance.

Short Answer

Expert verified
The rocket achieves a speed of approximately 165.71 m/s.

Step by step solution

01

Identify Problem Components

We need to find the speed of the rocket after being launched. The problem provides the total impulse and the mass of the rocket, while indicating that we should neglect gravity and air resistance effects.
02

Recall the Impulse-Momentum Theorem

The Impulse-Momentum Theorem states that the impulse experienced by an object is equal to the change in momentum of that object. The formula is: \[ F \Delta t = m \Delta v \]where \( F \Delta t \) is the impulse and \( m \Delta v \) is the change in momentum.
03

Substitute the Known Values

We are given that the impulse \( F \Delta t = 29.0 \mathrm{N} \cdot \mathrm{s} \) and the mass \( m = 0.175 \mathrm{kg} \). The initial velocity is zero since the rocket is at rest, so we can write:\[ 29.0 = 0.175 \times v \]
04

Solve for the Final Velocity

Rearrange the equation to solve for \( v \), the final velocity:\[ v = \frac{29.0}{0.175} \]Calculate:\[ v = 165.71 \mathrm{m/s} \]

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Key Concepts

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

Understanding Impulse
Impulse in physics refers to the change in momentum of an object when a force is applied over a period of time.
Essentially, it describes how much of an influence a force has on an object鈥檚 motion. The impulse experienced by an object is equal to the force multiplied by the time over which it is applied.
The formula for impulse is given by:
  • Impulse = Force \( \times \Delta t \)
Impulse has the same units as momentum, which are \( \text{N} \cdot \text{s} \) (newton-seconds) or \( \text{kg} \cdot \text{m/s} \) (kilograms m/s).
In the context of our rocket problem, the motor provides a total impulse of \( 29.0 \text{ N} \cdot \text{s} \).
This impulse is what causes the change in velocity of the rocket.
Exploring Momentum
Momentum is a fundamental concept in understanding how objects move and interact.
It's the product of an object's mass and its velocity. This means that an object with a larger mass or higher velocity will have more momentum.
The formula is given by:
  • Momentum = mass \( \times \) velocity or \( p = m \cdot v \)
Momentum is a vector quantity, meaning it has both magnitude and direction.
In our rocket scenario, initially, the rocket is at rest so the momentum is zero. After the motor provides an impulse, the momentum changes, propelling the rocket forward.
Velocity Calculation from Impulse and Momentum
Calculating velocity from impulse and momentum involves applying the Impulse-Momentum Theorem.
This states that the impulse delivered to an object results in a change in that object's momentum, expressed as:
  • Impulse = Change in Momentum or \( F\Delta t = m \Delta v \)
For our exercise, given that the impulse is \( 29.0 \text{ N} \cdot \text{s} \), and the rocket's mass is \( 0.175 \text{ kg} \), we use:
  • \( 29.0 = 0.175 \times v \)
  • Solve for \( v \), the rocket's final velocity \( v = \frac{29.0}{0.175} \) leading to \( v = 165.71 \text{ m/s} \)
So, the velocity of the rocket when launched from rest is approximately \( 165.71 \text{ m/s} \).
This calculation is straightforward due to the absence of external factors like gravity and air resistance.
Physics Problem-Solving Approach
When tackling physics problems, especially those involving impulse and momentum, having a systematic approach can be very helpful.
Here鈥檚 a simplified strategy for solving these problems:
  • Understand the problem. Identify what you need to find and what is given.
  • Recap relevant physics principles, such as the Impulse-Momentum Theorem in this exercise.
  • Apply formulas and substitute known values.
  • Calculate the unknowns, like the final velocity in our problem.
Always remember to review the assumptions made, such as neglecting air resistance or gravity in this scenario.
This helps in isolating the main physical concepts and delivering clear solutions.

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Most popular questions from this chapter

A \(40.0-\mathrm{kg}\) boy, riding a \(2.50-\mathrm{kg}\) skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's veloc- ity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

Two particles are moving along the \(x\) axis. Particle 1 has a mass \(m_{1}\) and a velocity \(v_{1}=+4.6 \mathrm{m} / \mathrm{s} .\) Particle 2 has a mass \(m_{2}\) and a velocity \(v_{2}=-6.1 \mathrm{m} / \mathrm{s} .\) The velocity of the center of mass of these two particles is zero. In other words, the center of mass of the particles remains stationary, even though each particle is moving. Find the ratio \(m_{1} / m_{2}\) of the masses of the particles.

For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the platform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that \(0.050 \mathrm{kg}\) of blood is pushed out of the heart with a velocity of \(+0.25 \mathrm{m} / \mathrm{s}\) and that the mass of the patient and platform is 85 kg. Assuming that the patient does not slip with respect to the platform, and that the patient and platform start from rest, determine the recoil velocity.

A golf ball strikes a hard, smooth floor at an angle of \(30.0^{\circ}\) and, as the drawing shows, rebounds at the same angle. The mass of the ball is \(0.047 \mathrm{kg},\) and its speed is \(45 \mathrm{m} / \mathrm{s}\) just before and after striking the floor. What is the magnitude of the impulse applied to the golf ball by the floor? (Hint: Note that only the vertical component of the ball's momentum changes during impact with the fl oor, and ignore the weight of the ball.)

A 5.00-kg ball, moving to the right at a velocity of \(+2.00 \mathrm{m} / \mathrm{s}\) on a frictionless table, collides head-on with a stationary \(7.50-\mathrm{kg}\) ball. Find the final velocities of the balls if the collision is (a) elastic and (b) completely inelastic.
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