Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 7: Problem 6

Consider two carts, of masses \(m\) and \(2 m,\) at rest on a frictionless air track. If you push the lower-mass cart for \(35 \mathrm{~cm}\) and then the other cart for the same distance and with the same force, which cart undergoes the larger change in momentum? a) The cart with mass \(m\) has the larger change. b) The cart with mass \(2 m\) has the larger change. c) The change in momentum is the same for both carts. d) It is impossible to tell from the information given.

Short Answer

Expert verified
Answer: b) The cart with mass 2m has the larger change.

Step by step solution

01

Recall the concept of momentum

Momentum is a vector quantity which is the product of mass and velocity of an object. Mathematically, momentum (p) is given by: p = m*v Where m is the mass of the object, and v is its velocity.
02

Recall the relation between force, mass, and acceleration:

According to Newton's second law of motion, the force (F) acting on an object is equal to the product of the mass (m) and the acceleration (a) of the object. Mathematically, F = m*a
03

Define the change in momentum

The change in momentum can be defined as the difference between the final momentum (pf) and the initial momentum (pi) of an object. Since both carts are initially at rest, their initial momentum is zero. Therefore, the change in momentum is equal to the final momentum. Mathematically, Δp = pf - pi
04

Calculate the acceleration of the carts

We know that both carts are pushed with the same force. Let's call this force F. Since the mass of the first cart is m and the mass of the second cart is 2m, we can calculate the acceleration of the first cart (a1) and the second cart (a2) using Newton's second law (F = ma) as follows: a1 = F/m a2 = F/(2m)
05

Calculate the final velocity of the carts

We know that both carts are pushed for the same distance (35 cm or 0.35 m). We can find the final velocity of each cart using the following relation between acceleration, initial velocity, final velocity, and distance (assuming constant acceleration): v^2 = u^2 + 2*a*s where v is the final velocity, u is the initial velocity (0 m/s for both carts), a is the acceleration, and s is the displacement. For cart 1: v1^2 = 0^2 + 2*a1*0.35 For cart 2: v2^2 = 0^2 + 2*a2*0.35
06

Compare the change in momentum

Now that we have the final velocities of the carts, we can calculate their change in momentum and compare them. Δp1 = m * v1 Δp2 = (2m) * v2 Since a1 = F/m and a2 = F/(2m), the final velocities for cart 1 and cart 2 will not be the same, with v1 being larger than v2. Since mass of cart 1 is m and mass of cart 2 is 2m, the change in momentum of cart 2 will be larger. The answer is: b) The cart with mass 2m has the larger change.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In a severe storm, \(1.00 \mathrm{~cm}\) of rain falls on a flat horizontal roof in \(30.0 \mathrm{~min}\). If the area of the roof is \(100 \mathrm{~m}^{2}\) and the terminal velocity of the rain is \(5.00 \mathrm{~m} / \mathrm{s}\), what is the average force exerted on the roof by the rain during the storm?

A skateboarder of mass \(35.0 \mathrm{~kg}\) is riding her \(3.50-\mathrm{kg}\) skateboard at a speed of \(5.00 \mathrm{~m} / \mathrm{s}\). She jumps backward off her skateboard, sending the skateboard forward at a speed of \(8.50 \mathrm{~m} / \mathrm{s}\). At what speed is the skateboarder moving when her feet hit the ground?

A 3.0 -kg ball of clay with a speed of \(21 \mathrm{~m} / \mathrm{s}\) is thrown against a wall and sticks to the wall. What is the magnitude of the impulse exerted on the ball?

Astronauts are playing baseball on the International Space Station. One astronaut with a mass of \(50.0 \mathrm{~kg}\), initially at rest, hits a baseball with a bat. The baseball was initially moving toward the astronaut at \(35.0 \mathrm{~m} / \mathrm{s},\) and after being hit, travels back in the same direction with a speed of \(45.0 \mathrm{~m} / \mathrm{s}\). The mass of a baseball is \(0.14 \mathrm{~kg}\). What is the recoil velocity of the astronaut?

A sled initially at rest has a mass of \(52.0 \mathrm{~kg}\), including all of its contents. A block with a mass of \(13.5 \mathrm{~kg}\) is ejected to the left at a speed of \(13.6 \mathrm{~m} / \mathrm{s} .\) What is the speed of the sled and the remaining contents?
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Circular Motion and Gravitation

Read Explanation

Fluids

Read Explanation

Space Physics

Read Explanation

Astrophysics

Read Explanation

Fields in Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free