Use the approximation that v⃗avg = p⃗/m for each time...

Use the approximation that $\vec{v}_{avg} = \vec{p}/m$ for each time step. A spring with a relaxed length of 25 cm and a stiffness of 10 N/m stands vertically on a table. A block of mass 85 g is attached to the top of the spring. You pull the block upward, stretching the spring until its length is now 29.4 cm, hold the block at rest for a moment, and then release it. Using a time step of 0.1 s, predict the position and momentum of the block at a time 0.2 s after you release the block. (Assume the + y direction is upward. Express your answers in vector form.) Part 1 (a) $\vec{r} = \langle \qquad \rangle $ m Part 2 (b) $\vec{p} = \langle \qquad \rangle $ kg\cdot m/s

Transcript

00:02 All right, looks like you have a conservation of energy problem here.

00:08 So let's list out what we know first.

00:11 So we know that the block is 1 .51 kilograms in mass.

00:19 It's moving to the right at a velocity of 4 .33 meters per second.

00:26 The track is frictionless and horizontal.

00:30 So we have no element of potential energy coming into play, gravitational potential energy anyways, and no loss energy due to friction.

00:40 Clides with a massless spring attached to a second block of mass 2.

00:47 So mass 2 is 2 .12 kilograms, and mass 2's initial velocity was negative 2 .15.

01:09 And then they give us a spring constant of 5 .5 by 5 .5.

01:14 Times 10 to the second dupons per meter.

01:28 All right, so we have that information.

01:30 Now let's look at what the questions are.

01:32 Determine the velocity of block two at the instant when block one is moving to the right with a velocity of three meters per second as showed in figure b.

01:45 And it looks like you have that correct.

01:47 So it looks like you use conservation momentum correctly.

01:50 So i'm not going to solve that part because it looks like you have that correct.

01:55 Already.

01:56 Notice it is a negative number.

01:58 That means that it's moving in the same direction as mass two now.

02:13 Okay, so let's just find the compression of the spring.

02:24 Are we supposed to find it at the point when they're at least three meters per second and 1 .2 or how much it was at its very maximum compression? i guess we're doing it at the moment when we have.

02:41 Okay, so i i think.

02:45 All right.

02:45 So this is what i'm going to interpret the question, is that we need to find the energy before the collision, which is the total kinetic energy initial.

03:02 And that needs to be equal to the kinetic energy after the collision in the scenario that they have here.

03:12 It says the term of the velocity at the incident when block one is moving to the right with the velocity of the plus.

03:17 Three meters per second.

03:21 So i'm going to count that as block one's v1f 3 .00 meters per second.

03:31 And you found the v2f, which is negative 1 .203 meters per second.

03:42 So we're looking at the final kinetic energy and whatever's not in that amount needs to be in the potential energy of the spring right now.

03:57 So we're using the conservation of mechanical energy here on the idea that whatever mechanical energy we had before the collision and the spring being constricted or compressed, that must equal the total mechanical energy afterwards, which means that something of it is now in the kinetic energy, but there's still compression in the spring.

04:23 So it must be, it must still be rebounding from this collision.

04:28 And so it's going to end up with more velocity in the end for each of these.

04:33 But as long as there's no outside forces on the system, and they're not sticking together, so we're going to assume this is a completely inelastic collision.

04:45 Doesn't specifically say that, but we're going to make that assumption, which means that that after the compression is finished, all of the final kinetic energy will equal the initial kinetic energy.

04:58 But when it's still compressed, some of it's being stored in the spray.

05:02 So let's see what we've got.

05:04 So we have some of the kinetic energy to start with is from the first object, m1b1i squared.

05:16 And then we also have some of the kinetic energy in the second object, m2, v2i squared.

05:23 So those are coming from their generic kinetic energy equation.

05:31 We've got two components to that kinetic energy.

05:35 Remember, energy is a scalar, so the negative sign is not going to be important here.

05:42 It's actually going to cancel out anyways when we square it.

05:47 And then we also have one half and one v1f squared plus one half and two.

06:00 V2 f squared and then we also have this little piece of spring potential energy which is one half k x squared and the x is the lone variable that we do not know so let's see what we can do here we've got one half times 1 .51 times 4 .33 squared plus one half times 2 .12 times a negative 2 .15 squared must equal 1 half times 1 .51 times its final velocity of 3, you know, 0 squared, plus 1 1 .2 times 2 .12 times negative 1 .20 squared squared.

07:30 A lot going on on this one, lots of numbers.

07:36 One half, and we're given the k value.

07:39 I'm just going to put $5 .55 to save some space here, x squared...

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