/*! 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 7 An object has a kinetic energy o... [FREE SOLUTION] | 91Ó°ÊÓ

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An object has a kinetic energy of \(275 \mathrm{~J}\) and a momentum of magnitude \(25.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). Find the speed and mass of the object.

Short Answer

Expert verified
The mass of the object is approximately 5 kg and its speed is approximately 5 m/s.

Step by step solution

01

Write down the given quantities

From the problem, we have two known quantities - kinetic energy (KE) = 275 J and momentum (p) = 25.0 kg.m/s.
02

Write down the formulas for kinetic energy and momentum

The kinetic energy of an object is given by the formula \( KE = \frac{1}{2}mv^2 \) and the momentum by \( p = mv \).
03

Substitute the known quantities into the formulas

Substitute the known quantities into the formulas to create a system of equations. This gives: \( 275 = \frac{1}{2}mv^2 \) and \( 25.0 = mv \) .
04

Solve for one of the unknowns

To isolate one of the unknowns, take the momentum equation \( p = mv \), and solve for \( v \) to get: \( v = \frac{p}{m} \).
05

Substitute the expression for velocity

Substitute the expression for velocity obtained in step 4 into the kinetic energy equation. This gives: \( 275 = \frac{1}{2}m\left(\frac{p}{m}\right)^2 \) . Simplify the equation to find the value of the mass. The mass \( m \) obtained is approximately 5 kg.
06

Substitute the value of mass into the momentum equation

The mass obtained in step 5 can be substituted into the momentum equation \( 25.0 = mv \) to find the speed. The speed \( v \) is approximately 5 m/s.

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Key Concepts

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

Momentum
In physics, momentum is a concept that is crucial for understanding motion and collision mechanics. You can think of momentum as a measure of how difficult it is to stop a moving object. The more momentum an object has, the more force is needed to change its direction or to stop it. Momentum is calculated using the formula:
  • \( p = mv \)
Here, \( p \) represents momentum, \( m \) is the mass, and \( v \) is the velocity.
Momentum is a vector quantity, which means it has both magnitude and direction. This is why two objects with the same speed but different directions have different momenta. When working with problems involving momentum, it's important to keep track of these vectors.
In the exercise, we know the momentum of the object is 25.0 kg·m/s. With momentum and kinetic energy both given, we can use these values to find the other parameters like mass and velocity.
Mass
Mass is a fundamental concept in physics. It represents the amount of matter in an object and is usually measured in kilograms. Mass is an intrinsic property of an object, which means it doesn't change regardless of where the object is in the universe.
In the context of the exercise, we use mass in the formulas for both kinetic energy and momentum. The relations are:
  • Kinetic energy: \( KE = \frac{1}{2}mv^2 \)
  • Momentum: \( p = mv \)
Given the momentum and kinetic energy values, we can solve these equations to isolate the mass (\( m \)). Through the provided calculations, we find that the mass of the object is approximately 5 kg.
Understanding mass helps in predicting how an object will behave when subjected to forces, making it a vital parameter in physics calculations.
Velocity
Velocity is another key concept in physics that refers to the speed of an object in a specific direction. Unlike speed, which is a scalar quantity, velocity is a vector, as it incorporates both magnitude and direction.
The formula for velocity in the context of momentum is:
  • \( p = mv \) which rearranges to \( v = \frac{p}{m} \)
Velocity plays an integral role in determining kinetic energy as well since it directly affects the energy an object possesses due to its motion.
In the exercise, once we determined the mass to be 5 kg, we could calculate the velocity of the object by substituting the mass back into the momentum equation. We find that the velocity is approximately 5 m/s.
Understanding velocity and its distinction from speed is crucial for solving problems related to motion and energy.

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Most popular questions from this chapter

A \(0.030-\mathrm{kg}\) bullet is fired vertically at \(200 \mathrm{~m} / \mathrm{s}\) into a \(0.15\) \(\mathrm{kg}\) baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?

ecp Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater B along a horizontal direction. Identify (a) the horizontal forces acting on \(\mathrm{A}\) and (b) those acting on \(\mathrm{B}\). (c) Which force is greater, the force on \(\mathrm{A}\) or the force on B? (d) Can conservation of momentum be used for the system of \(A\) and \(B\) ? Defend your answer. (e) If \(A\) has a mass of \(0.900\) times that of \(\mathrm{B}\), and \(\mathrm{B}\) begins to move away with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\), find the speed of \(\mathrm{A}\).

A \(60-\mathrm{kg}\) soccer player jumps vertically upwards and heads the \(0.45\) -kg ball as it is descending vertically with a speedof \(25 \mathrm{~m} / \mathrm{s}\). If the player was moving upward with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? If the ball is in contact with the player's head for \(20 \mathrm{~ms}\), what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

ecp (a) A car traveling due east strikes a car traveling due north at an intersection, and the two move together as a unit. A property owner on the southeast corner of the intersection claims that his fence was torn down in the collision. Should he be awarded damages by the insurance company? Defend your answer. (b) Let the eastward-moving car have a mass of \(1300 \mathrm{~kg}\) and a speed of \(30.0 \mathrm{~km} / \mathrm{h}\) and the northward-moving car a mass of \(1100 \mathrm{~kg}\) and a speed of \(20.0 \mathrm{~km} / \mathrm{h}\). Find the velocity after the collision. Are the results consistent with your answer to part (a)?

A \(12.0-\mathrm{g}\) bullet is fired horizontally into a \(100-\mathrm{g}\) wooden block initially at rest on a horizontal surface. After impact, the block slides \(7.5 \mathrm{~m}\) before coming to rest. If the coefficient of kinetic friction between block and surface is \(0.650\), what was the speed of the bullet immediately before impact?

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