a bullet with mass m 01 kg grams hits a ballistic pendulum with length l 3 meters and mass m 2 k

A. While the bullet is embedding into the block, energy is not conserved. This is because some o

A bullet with mass m 01 kg grams hits a ballistic pendulum...

A bullet with mass m = 0.1 kg hits a ballistic pendulum with length L = 3 meters and mass M = 2 kg and lodges in it. When the bullet hits the pendulum, it swings up from the equilibrium position and reaches an angle of 15 degrees at its maximum. In this problem, you will determine the bullet's velocity. First, I want to ask you a couple of questions: While the bullet is embedding into the block, is energy conserved? Why or why not? B. While the bullet is embedding into the block, is momentum conserved? Why or why not? C. When the pendulum swings up to its maximum angle, is energy conserved? Why or why not? D. When the pendulum swings up to its maximum angle, is momentum conserved? Why or why not? E. Find the velocity of the bullet.

Transcript

00:01 In this problem, we're looking at both conservation of momentum and conservation of energy.

00:06 We have two states.

00:07 We have the first state, which is before the collision, where i have a pendulum.

00:16 And that pendulum has a mass.

00:19 And then there's a little bullet speeding towards the pendulum with a, i'm actually going to denote this with a mass capital m.

00:29 And i'm going to give this little mass m.

00:32 And then after the collision, state.

00:35 To the bullet and the pendulum have collided.

00:42 The pendulum is now at a 15 degree angle.

00:48 Okay, so there's the pendulum and the block embedded together, and it's kind of kind of gone up to a 15 degree angle and stopped.

00:58 The other thing we need to note is that the length of this pendulum is three meters.

01:04 We're also given values for big m the mass of the block to be 2 kilogram and little m the mass of the bullet to be 0 .1 kilograms.

01:16 And that means that the mass over here, big m plus little m is 2 .1 kilograms.

01:24 We want to figure out what the velocity of the bullet is.

01:28 We know the velocity of the block is equal to 0.

01:32 And we know that the velocity over here is zero as well.

01:39 Okay, once it stops, it's not moving.

01:42 Okay, so before we start setting up this problem, let's answer a couple of key questions.

01:47 The first question asks if energy is conserved, no, energy is not conserved because this is an inelastic collision, and energy is not conserved in an inelastic collision.

02:05 We actually lose energy to heat and things like that.

02:08 So on a grand scale, yes, it is conserved, but actually just in the collision itself, no energy does not appear to be conserved.

02:16 Yes, momentum is always conserved in a collision.

02:24 So both of these are dealing with part one.

02:26 Okay.

02:27 Then we're asked, is energy conserved in part two? and yeah, that's true.

02:32 As that pendulum moves, it starts moving, has kinetic energy that gets converted to potential energy.

02:40 That is energy conservation.

02:44 And then no, momentum is not conserved in this case because we're not talking about a collision of any kind.

02:52 The collision has already happened.

02:53 So what the problem is asking us to do is think about these two points, and that's pointing us towards how we want to set up our equation.

03:01 So the next thing we want to do is we want to set up equations for part one and for part two, which are going to rely on conservation of momentum and conservation of energy respectively.

03:14 So let's look at part one.

03:17 This is our momentum conservation.

03:25 Okay.

03:26 Momentum conservation is going to tell me that the mass of the bullet times the velocity of the bullet, that's ultimately what we're looking for, plus the mass of the block times the velocity of the block.

03:39 But hey, y 'all, oh, that's zero, is equal to the combined mass, mb plus m of the block, times the total velocity of the both of that...

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