Download the App!

Get 24/7 study help with the Numerade app for iOS and Android! Enter your email for an invite.

Two blocks of mass $m_{\mathrm{A}}$ and $m_{\mathrm{B}},$ resting on a frictionless table, are connected by a stretched spring and then released (Fig. $51 ) .(a)$ Is there a net external force on the system? (b) Determine the ratio of their speeds, $v_{A} / v_{B}$ . (c) What is the ratio of their kinetic energies? (d) Describe the motion of the $\mathrm{CM}$ of this system. (e) How would the presence of friction alter the above results?

Get the answer to your homework problem.

Try Numerade free for 7 days

Like

Report

a) No external forceb) $m_{B} / m_{A}$c) $m_{B} / m_{A}$d) Remains at reste) With friction present, there could be a net external force on the system, because the forces of friction on the two masses would not necessarily be equal in magnitude. If the two friction forces are not equal in magnitude, the ratios found above would not be valid. Likewise, the center of mass would not necessarily be at rest with friction present.

Physics 101 Mechanics

Chapter 9

Linear Momentum

Motion Along a Straight Line

Kinetic Energy

Potential Energy

Energy Conservation

Moment, Impulse, and Collisions

University of Michigan - Ann Arbor

University of Washington

University of Sheffield

University of Winnipeg

Lectures

04:05

In physics, a conservative force is a force that is path-independent, meaning that the total work done along any path in the field is the same. In other words, the work is independent of the path taken. The only force considered in classical physics to be conservative is gravitation.

03:47

In physics, the kinetic energy of an object is the energy which it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes. The same amount of work is done by the body in decelerating from its current speed to a state of rest. The kinetic energy of a rotating object is the sum of the kinetic energies of the object's parts.

03:17

Two blocks of mass $m_A$ a…

03:34

'Two blocks of mass m…

01:59

When two blocks $A$ and $B…

01:53

05:17

The mass m = 5.5 kg restin…

01:28

Two blocks lie on each oth…

04:00

A 15.0 -kg block rests on …

08:39

Block $A$ in Fig. E8.24 ha…

00:56

Two blocks of masses $M$ a…

03:01

A block of mass M is at re…

07:49

Block A in $\textbf{Fig. E…

03:55

A block of mass $M$ is at …

All right. So for port A, um uh, is there no force on the system? The answer isn't No. The net force on the system is zero. There's no external force acting on the system. Um and that's because that spring forces internal to the systems, the system itself is an equilibrium and partly BB. We use momentum Conservation National momentum recalls final momentum eso initially of zero momentum, of course. And then you have Matic Walt, maybe a minus and B B B. Because the blocks are moving in opposite directions. And so v a over B will just be m b over Emma in part C, we want ratio of kinetic energy is so okay over Kay Beauty is just 1/2 and a V a squared over on half and b u d squared on the one has canceled. So m et over m b times V a over Vehbi quantity squared is equal to m a over and be times whatever we're plays video over be from there. So that's m B over M a. Ah, quantity squared. And so we're left to it and be over m a as ratio in part, see same ratio of kinetic energies as velocities and then part d. What happens to the center of mass of the system so initially? Well, weapons, too, though initially, center mass of the system is zero. Uh, but we also have that there is no net force of the system. So therefore, finally, senator less system is also zero. Um, and so the senator massive system does not change, essentially in part B. What happens if there's a force of friction? So the force of friction there will be a net external force. That's not zero. Ah. So the ratios in part, uh, parts B and C are not valid anymore. They're not valid. Ah, And then, of course, since the Net force is not zero, the center of mass is also mom zero, and that does change.

View More Answers From This Book

Find Another Textbook