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Blocks $A$ (mass 2.00 $\mathrm{kg} )$ and $B(\mathrm{mass} 10.00 \mathrm{kg})$ move on a frictionless, horizontal surface. Initially, block $B$ is at rest and block $A$ is moving toward it at 2.00 $\mathrm{m} / \mathrm{s}$ . The blocks are equipped with ideal spring bumpers, as in Example 8.10 . The collision is head-on, so all motion before and after the collision is along a straight line. (a) Find the maximum energy stored in the spring bumpers and the velocity of each block at that time. (b) Find thevelocity of each block after they have moved apart.

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(a) $U_b = 3.33 \mathrm{J}$ and $v’=0.333 \mathrm{m} / \mathrm{s}$(b) $v_b=+0.667 \mathrm{m} / \mathrm{s}$ and $v_a=-1.34 \mathrm{m} / \mathrm{s} $

Physics 101 Mechanics

Chapter 8

Momentum, Impulse, and Collisions

Moment, Impulse, and Collisions

Cornell University

Rutgers, The State University of New Jersey

University of Washington

Hope College

Lectures

04:30

In classical mechanics, impulse is the integral of a force, F, over the time interval, t, for which it acts. In the case of a constant force, the resulting change in momentum is equal to the force itself, and the impulse is the change in momentum divided by the time during which the force acts. Impulse applied to an object produces an equivalent force to that of the object's mass multiplied by its velocity. In an inertial reference frame, an object that has no net force on it will continue at a constant velocity forever. In classical mechanics, the change in an object's motion, due to a force applied, is called its acceleration. The SI unit of measure for impulse is the newton second.

03:30

In physics, impulse is the integral of a force, F, over the time interval, t, for which it acts. Given a force, F, applied for a time, t, the resulting change in momentum, p, is equal to the impulse, I. Impulse applied to a mass, m, is also equal to the change in the object's kinetic energy, T, as a result of the force acting on it.

07:43

Blocks $A($ mass 2.00 $\ma…

04:05

Blocks $A$ (mass $2.00 \ma…

09:56

Blocks $A$ (mass 2.00 kg) …

04:33

Blocks A (mass 2.00 kg) an…

04:57

03:28

15:53

04:10

Two Blocks and a Spring A …

09:15

A $100 \mathrm{g}$ block o…

02:42

Spring Attached to Wall A …

01:06

Two blocks 1 and 2 of mass…

01:59

A $15.0 \mathrm{~kg}$ bloc…

in this question. We have two blocks block A with a mass of 2 kg, is moving to the right with the velocity off 2 m per second. Block B, with a mass off 10 kg, is just standing still and both blocks are keep it with ideal spring bumpers. In the first item off this question, we have to calculate what is the maximum energy stored in the bumpers and then what is the velocity off those blocks? At that moment, we begin by noting the following. When does bumpers are compressed to the maximum? What happens is that the velocity off block A is equal to the velocity off block B and this velocity I'm calling the prime. Then, knowing that we can use the principal off mo mentum conservation, the principal off momentum conservation tells us that the net momentum is conserved before and after a given situation. So the net momentum pew net before is, of course, of the net momentum pew net after then. By before I mean the beginning of the situation where Onley block a is moving to the right. The reform all the momentum is doing to block A, which has a momentum given by M a. Times v a. Then when that spring is compressed to the maximum, both blocks are moving to the same direction with the same velocity. The reform pew net after is given by the mass off block eight times the prime, plus the mass off block B times the prime. Now all we have to do to discover V prime is so this equation for V prime, which is not a difficult task. All you have to do is factor v prime on the right hand side and then solve for V prime straightforwardly just like that. So v Prime is given by M a Times va divided by m A plus and be Now we plug in the values that were given by the problem and they get the following m A. Is equals to choose. VA is equals two plus two. Given that we choose these reference frame where everything that points to the right is pointing to the positive direction and then we divided by M A, which is true plus and B which is stand. These results in four divided by 12 and this is 1/3, which is approximately 0.333 m per second knows that I have rounded this result to three significant figures since we always had three significant figures throughout the problem. Then when the springs compressed to the maximum, the velocity off both blocks is 0.333 m per second. Now we have to evaluate what is the maximum energy storing the bumpers. For that, we have to remember about conservation off energy. So during an elastic collision like this one, energy is conserved. This means that we can use the principal off energy conservation to evaluate what is the potential energy stored in the bumpers. The principal off energy conservation tells us that the mechanical energy E m before is equal to the mechanical energy after and by before. I mean, when Onley block A is moving to the right When Onley block is moving to the right, all the energy is actually kinetic energy off block A. So before we only have the kinetic energy off block A, which is given by m a times va squared, divided by true and after what we have is two blocks that are moving. So we have the kinetic energy off block eight and eight times V prime squared, divided by two. We have the kinetic energy off block B and B times V prime divided by two. And we also have that potential energy stored in the bumpers you be then, as we want to discover you be we solve this equation for you Be on This is not difficult. We get the following you be is equals to m a va squared divided by two minus m a v prime squared divided by true minus M b v Prime squared divided by true Then now we plug in the values that were given by the problem and the value off the prime that we had calculated before. So I m a is equal to two v a is true. Then we divide by two. Now we have a again which is true v prime, which is 0.333 squared, divided by two minus m B, which is 10 times v prime 0.333 squared, divided by two on. By solving this, we got ah potential energy off approximately 3.34 jewels. And this is the answer to the first item off this question and now we go to the second item in the second item, we have to evaluate what is the velocity off block A after they separate after the collision. And what is the velocity off block B? Also after the collision. For that we can use these equations that tells us what are the velocities off blocks A and B after the collision provided that block B begins at rest, which is our case. Then, by applying these equations, we get the following V. A prime is given by the mass off block A, which is true minus the mask off block B, which is 10 divided by the mass off block A plus the mass off block B. And this multiplies the initial velocity off block A. This gives us minus eight divided by 12 times true which is minus 16 divided by 12 on this is approximately minus 1.33 m per second. So this is the velocity off block A after the collision. Now for VB prime, we have the following vb prime is equals to two times the mass off block A, which is true, divided by the mass off block a which is true plus the mass off block B times the initial velocity off Block A. This results in eight divided by 12 which is approximately 0.667 m per second, and this is the answer to the second item off this question.

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