Two children are playing a game in which they try to hit a...

Key Concepts

Conservation of Energy

This principle states that energy in an isolated system remains constant, permitting the conversion of one form of energy to another. In mechanics, it allows us to equate the potential energy stored in a compressed spring to the kinetic energy of an object once released, thus determining the initial speed of the projectile provided there are negligible losses.

Elastic Potential Energy

Elastic potential energy is the energy stored in an object when it is deformed, such as when a spring is compressed. It is given by the formula ½ k x², where k is the spring constant and x is the displacement from equilibrium. This stored energy is crucial in launching projectiles in systems that rely on spring mechanisms.

Projectile Motion

Projectile motion describes the motion of an object that is launched into the air and moves under the influence of gravity alone. This concept involves analyzing the horizontal displacement and the vertical motion separately, assuming no air resistance, and applying kinematic equations to predict the trajectory and impact point.

Kinematic Equations for Free-Fall

Kinematic equations provide a relationship between displacement, velocity, acceleration, and time. In the context of free-fall under gravity, these equations help determine the time taken for an object to fall a given vertical distance, which is an essential component in calculating the horizontal distance traveled by a projectile.

Transcript

00:01 Okay guys, now let's do problem 36 from chapter 8.

00:05 And this is once again a combination of problems.

00:08 There's kinematics involved and there is conservation of energy.

00:13 So it says that two children are playing a game in which they try to hit a small box on the floor with a marble fire from a spring loaded gun that is mounted on the table.

00:22 The target box is horizontally distance.

00:25 Oh, the target box's horizontal distance is 2 .2 meters from the edge of the table.

00:31 And so in figure 4 .8, okay, bobby compresses this gun spring 1 .1 centimeters from the center of the marble.

00:41 He compresses the spring 1 .1 centimeters, but the marble falls 27 centimeters short of the box.

00:48 And they want to know how far this girl rhoda should compress the spring to basically hit the box, which is 2 .2 meters away.

00:57 Well, this is obviously a kinematics problem here because they have this box thing.

01:06 Okay, and there's a spring inside this box.

01:11 And there is a box.

01:13 There is another box that is a certain distance away.

01:17 Okay, certain distance, d.

01:20 And we know what d is because d is equal to 2 .20 meters.

01:25 Now, this entire problem is very interesting.

01:28 All we need to do is simply work backwards and understand the principles involved here.

01:34 All right.

01:35 So let's start.

01:36 In order to solve this, what do we know? well, we know that, you know, if you stick a ball in here and it fires out, right? there's obviously conservation of energy occurring.

01:49 Okay? and we know that there is obviously potential energy of the spring, right? because the spring is being compressed in order to fire it out.

01:59 And when it fires out, there is kinetic energy.

02:03 And so we want to know basically what this like velocity is.

02:09 What is the velocity that this kid is using? you know, what velocity did this ball fly out? and with that velocity, we can solve for the time.

02:23 We can solve for the time required.

02:26 And when we solve for the time required, we can find the velocity needed and that we can solve for how far this spring has to compress in order for it to reach that velocity.

02:38 So a lot of working backwards here.

02:39 We are missing a lot of information.

02:41 We're like, what is k? right.

02:45 And what is k? we know what x is.

02:49 We don't know what a v is at the moment.

02:52 But we'll get to this.

02:54 But first, let's recognize that you can cancel some stuff out.

02:57 So we can cancel out the one -halfs, the one -half.

03:00 We know what delta x is in this problem.

03:02 Because when this boy, when this boy bobby first compresses it, he stretches it 1 .1 centimeters, which is 0 .01 meters.

03:15 Okay? so let's keep that in mind.

03:17 And 0 .01 meters, 0 .01 meters squared is 1 .21 times 10 to the negative 4.

03:28 Oh, man, that's a lot of write out.

03:30 All right.

03:30 1 .21 times 10 to the negative 4 meters times x is equal to whatever the mass is.

03:41 We don't even know what the mass of this thing is.

03:45 Mass times v final squared.

03:50 What is the mass? why does it not give the mass in this problem? well, anyway, we know that we can find the velocity somehow.

04:00 So i don't even think we need the mass.

04:03 Okay, so let's find the velocity.

04:05 How do we do that? well, we know that there's a kinematics equation, right? but first let's explain this.

04:13 When the ball shoots out, it's at a certain horizontal velocity, and it travels a certain distance d, and in a certain amount of time, it travels that distance d.

04:25 So we know that when it reaches the ground, there is a final velocity of zero, but we don't know what the initial velocity is.

04:32 So this is very interesting because there's a kinematics equation that deals with this, right? and isn't that kinematics equation, v final squared is equal to v initial squared plus two? 2a delta d where we recognize that v final is zero and we recognize that a is actually g because it's gravity acting downwards.

04:56 So it's 2g delta d is equal to v initial square.

05:00 And so our v initial is equal to the square root of 2g delta d.

05:05 And all we have to do simply substitute in what we know.

05:10 We know that bobby was 27 centimeters short of the 2 .20 meters.

05:18 So we know that we know that he was 2 .20 meters minus 0 .27 meters is equal to what? 2 .20 minus 0 .27 is equal...

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