Problem 2: A 200 gram hockey puck is launched up a metal...

Problem 2: A 200 gram hockey puck is launched up a metal ramp that is inclined at a 45° angle. The coefficients of static and kinetic friction between the hockey puck and the metal ramp are ?s = 1.60 and ?k = 1.50, respectively. The puck's initial speed is 6 m/s. (a) The vertical height reached by the puck above its starting point is .73 (b) The net work done on the puck is Problem 3: A particle with mass m1=2 kg is located at x=0 while a particle with mass m2=128 kg is located at x=200 m along the x axis. Somewhere between them is a point where the gravitational force of m1 acting on a mass m0=1 kg is canceled by the gravitational force of m2 acting on that mass. (a) What is the coordinate x of this point? (b) Find the ratio (in terms of integer numbers) of the magnitude of the net gravitational force on a mass m0=1kg located at (x=100, y=0) relative to the magnitude of the net gravitational force on the same mass m0 when located at (x=300, y=0) Problem 4: During a snowball fight two balls with respective masses of 1.6 and 0.4 kg are thrown in such a manner that they meet with their velocities forming a 30° angle. After the collision, the two snowballs combine to form a single mass. The speed of each snowball just before the collision is 15 m/s. (a) The speed of the combined snowball immediately after collision is (b) The kinetic energy of the combined snowball (with mass 2 kg) is Problem 5: A 2.0 kg block, lying on a smooth table, is connected to another 5.0 kg block by means of a weightless cord passing over, without slipping, a cylindrical pulley of 2.0 kg mass and 10.0 cm radius placed at the edge of the table. The system is released from rest at t = 0. Consider g=9.8 m/s². (a) In 0.3s the 5.0 kg block will drop by (b) The tension in the part of cord connected horizontally to the 2.0 kg block is

Transcript

00:03 All right, looks like you have quite a few questions on your, i assume this is a review sheet of the things that you've talked about over the course of this physics semester.

00:15 Because they're all different concepts here.

00:17 So i tried to set them all up for you so that you can practice this.

00:23 Question number two is all about energy without with the non -conservative, i should just be saying with the non -conservative force.

00:37 Let me change that for you with.

00:45 So the non -conservative force here is the friction.

00:48 So the amount of kinetic energy to start with must equal the amount of potential energy and the loss of energy due to friction.

00:58 So we can find the work done by friction using this equation.

01:02 You have all the numbers necessary to do that.

01:05 So that's an example of a good problem where we would still use that conservation.

01:10 Of energy idea with the idea that there's some work being done by a non -conservative force, it's taking some of that energy, but it still has to go somewhere.

01:21 It doesn't just disappear, so we can solve for that.

01:24 And that's what the work due to friction is equal to the energy lost from the system.

01:32 So that's number two.

01:34 Number three is a gravitational attraction problem.

01:39 So i have that one set up using newton's law of gravity.

01:45 So we have three masses that are located on this x plane here.

01:50 And the difference between m1 and m2 is 200 meters.

01:54 So the difference between m1 and m0, i'm calling x.

01:58 And that means that the difference between m0 and m2 is 200 minus x.

02:07 And so we know that it said in the problem that the two gravitational attractions, need to cancel each other.

02:15 So i took newton's law of gravity, fg equals g, m1, m2, divided by r squared, and i made them equal to each other.

02:24 The attraction from m1 to m0 is on the left side, and the attraction from m0 to m2 is on the right side.

02:34 Those two have to equal each other because they're canceling each other out.

02:39 So that's the two equations there.

02:41 You have all the information to solve for x.

02:44 Remember capital g is not 9 .8.

02:48 It's 6 .67 times 10 to the negative 11th, as long as we're using kilograms and meters in our measurements.

03:01 So that's how to set up number three to solve for x.

03:06 And then number four is a two -dimensional inelastic collision.

03:12 So this is a conservation momentum.

03:15 Anytime you have conservation momentum in two dimensions, you have to solve it in two parts, the x component and the y component.

03:25 So i made a drawing here, and the easiest way to do this is to create one of your two objects going in a only one component.

03:35 So that's what i did here.

03:36 I made m1 have a momentum that's in only the x component.

03:42 That way i don't have to do as much trig.

03:45 If i made them both in the, both in two dimensions, then i would have twice as much trig to do.

03:53 So on the x component, i all, all i have to do is take m1 v1x, which i know, that's the velocity i already known.

04:03 But m2v2x, i'm going to take, well, what's the x component of that? that is going to be the adjacent side in this scenario.

04:17 So that would you would need to take to find b2x, you would need to take the hypotenuse, which i believe they said was 15.

04:26 So i'm going to take a cosine of 30.

04:30 So because it's an adjacent side.

04:32 So cosine is what you would use there.

04:34 Cosine 30 is equal to the adjacent side, which you don't know divided by the hypotenuse, which i believe was the total velocity is 15.

04:46 So once you have that, you can find what the bx is.

04:52 And then the same with y component, but the nice thing is, is that whatever the y component's momentum is, whatever that's velocity is, which would be the sine 30 of the opposite over 15, that would tell us that we all, what what we know for total momentum, so all we have to do is do a little bit less, less, less trig there than the x component.

05:18 So y component and x component, i made a little triangle over here to give you an idea of what do you have to do next.

05:26 The v value, the amount of velocity, which is what's asked for, you need to use the pythagorean theorem once you have the y and the x component of velocity.

05:37 B, y, square plus vx squared square root and will tell you b.

05:41 And then use tangent.

05:43 Tangent is the opposite over adjacent.

05:45 So v .y over v .x inverse tangent will tell you the angle that this collision is now going at...

Question

Please give Ace some feedback