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Example #1: Elastic Collisions

**Transcript**

Hey guys, so let's check out this example of elastic collisions, now remember elastic collisions are special in that 1) kinetic energy is conserved so K final=K initial and 2) we can use an extra equation which is V1 initial+V1 final=V2 initial+V2 final, I want to talk about the situation real quick we're not actually going to use this is conceptual, this is what we're actually going to use as an extra equation and by the way the second question comes from the first equation this is just a useful version of the first one, OK? I want to point out something real quick this is different from this equation because first of all there are no masses but the other difference is that here you have initial 1, initial 2, final 1, final 2, here you have initial final, initial final So here it's 1,2 1,2 here's 11,22 so it's a little but different you got to figure out a way to remember that, so 1, 2, 1, 2 and 1, 1 2, 2, OK? So let's do this, 2 objects of the same mass undergone an elastic head on collision, the reason I'm able to use the second equation in addition to the first equation is because it told me here that this is elastic, head on just means that they're going towards each other like this, they have the same mass it doesn't tell me what the mass is so I'm just going to call it M for now, object A from the left (so this has to be A and therefore this is B) has speed of 3 and object B has speed of 4, remember opposite directions opposite signs +3-4 please don't forget the negative 4 or you're going to get the wrong answer, calculator final velocities, magnitude and direction, magnitude means that number and direction means positive or negative, OK? How are we going to do this? it's a collision conservation of momentum so M1 V1 or in this case Ma Va+Mb Vb both initial=Ma.Va+MbVb final, I know the masses......Actually I don't know the masses but they're all M so what's going to happen is whenever the 2 masses are the same you're going to be able to cancel it, right? So, whether it's M and M or if it's 10 and 10 whatever if the numbers are the same you can always cancel it from this equation, Va initial is right there=+3, Vb initial is -4. Va final I don't have and we're looking for it and Vb final I don't have and we're looking for it, so what we ask here could be well they just didn't give me enough information? But they did instead of giving one of these two velocities they replaced that with the information that this is elastic so it's a give and take, I tell you the type of collision but then I give you one less number for you to be able to solve this problem, OK? So instead of giving you a number I told you that it was elastic and you can still solve it but this is 2 unknowns with just one equation which means you can't solve it, what do you do? Well you need another equation so the other question you are going to use is this one right here, OK? So, you're going to use the equation, the special equation for elastic collision so the V1 initial + V1 final= V2 initial + V2 final, by the way there are a few different versions of this equation some of them have like negatives in them what not, I moved things and I showed it to you this way because I think it's the easiest one to remember but you might see it slightly different, alright? So let's plug in the numbers, the initial velocity here...The initial velocity of 1=+3, the final velocity of 1 I don't have it and this one is a -4 and this one I don't have it, so now that I have two equations I can actually solve this by combining the two equations so the first thing I'm going to do is I'm going to combine the numbers and put the variables in the same sides, so 3-4=-1 Va final+Vb final, there's nothing else I can do here for now, let's do the same thing here let's move this number over here so that I have numbers on the left and letters in the right, so the 4 goes over here I got a 7 equals and then this guy goes over here and then it's going to a negative, negative this is A and B actually, right? -Vaf +Vbf, now notice that these two equations have the same basic structure, number an A and the B, a number the A and the B, OK? So, what I can do now is I can move this guy over here and we're going to combine these two equations and the easiest way to do this is by adding the equations so let's stack them up on top of each other. You can add equations remember that and the way that you add equations is everything to the left just as a quick reminder everything to the left of the equals sign gets added so -1+7=+6plus and everything to the right of the equals sign gets added so I have Va+-Va so Va+-Va just cancels equals Vb+Vb that's 2 Vb, 2Vbf which means Vbf is 6/2=3, +3, Vbf=+3. Now that I got one of the two velocities I can simply get that number and plug it into either one of the 2 original equations, Ok? To find Vaf, I'm going to plug into this one here just because that equation the Va is already positive, OK? Let's move this over here and I have a -1 equals... Whoops Vaf+Vbf, -1 Vbf is 3, it's going to go to the left as -3, Vaf=-4 meters per second, OK? This is positive so it's going to the right so B after the collision is going to go to the right with 3 and A after the collision is going to go to the left with 4, OK? You might have noticed that the numbers flipped, this guy was a 3 ended up with a 4, this guy is a 4 and ended up with a 3 that's what happens if you have elastic collisions but only happens if you have exactly the same mass, the velocity simply flip, OK? So that's it for this one, elastic collisions are a little bit more complicated a little bit more work because you do have fewer numbers that are going to be given to you and you do have to use two equations to solve them, there is the big equation, the big momentum equation and you got to use the extra elastic collision equation, so that's it let me know if you have any questions.

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Example #1: Elastic Collisions

Blocks A and B are moving on a horizontal frictionless air track. Block A has mass 3.0 kg and block B has mass 6.0 kg. Initially block A is moving to the right at 6.0 m/s and block B is moving to the left at 8.0 m/s. The blocks collide and after the collision block A is moving to the left at 4.0 m/s. What is the velocity of block B after the collision?
A) 3.0 m/s to the left
B) 3.0 m/s o the right
C) 8.6 m/s to the left
D) 8.6 m/s to the right
E) 13.0 m/s to the left
F) 13.0 m/s to the right
G) None of the above answers

A 5 kg and a 3 kg mass collide head on, with the 5 kg mass moving at an initial speed of 15 m/s and the 3 kg mass moving at an initial speed of 10 m/s. Answer the following questions:
a) If the collision is completely inelastic, what is the final speed and direction of each mass?
b) If the collision is elastic, what is the final speed and direction of each mass?

A 46.0 g marble moving at 2.20 m/s strikes a 26.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.(a) What is the speed of 46.0 g marble immediately after the collision?(b) What is the speed of 26.0 g marble immediately after the collision?

In the “slingshot effect,” the transfer of energy in an elastic collision is used to boost the energy of a space probe so that it can escape from the solar system. All speeds are relative to an inertial frame in which the center of the sun remains at rest. A space probe moves at 6.4 km/s toward Saturn, which is moving at 6.4 km/s toward the probe. Because of the gravitational attraction between Saturn and the probe, the probe swings around Saturn and heads back (nearly) in the opposite direction with speed vf. By what factor is the kinetic energy of the probe increased?
1. 16.2591
2. 22.1363
3. 9.14008
4. 4.80263
5. 6.404
6. 7.94215
7. 3.7032
8. 9.0
9. 3.61
10. 9.33331

A projectile of mass m 1 moving with a speed v1 in the +x direction strikes a stationary target of mass m2 = 2m1 head-on in an elastic collision. Find the final velocity of the projectile m1.
Hint: You can use the energy and momentum principles.
1. -1/3 v1
2. -1/4 v1
3. -1/5 v1
4. 1/3 v1
5. -1/2 v1
6. v1
7. -5 v1
8. 3 v1

A 0.060-kg tennis ball, moving with a speed of 5.1 m/s, has a head-on collision with a 9.5×10−2-kg ball initially moving in the same direction at a speed of 2.8 m/s. Assuming a perfectly elastic collision, determine the speed and direction of each ball after the collision.

An 54.5 kg object moving to the right at 55.9 cm/s overtakes and collides elastically with a second 35.7 kg object moving in the same direction at 39 cm/s. Find the velocity of the second object after the collision.1. 39.26342. 56.50273. 38.82144. 55.32435. 59.42246. 39.98137. 58.31228. 57.21069. 41.241410. 48.0473

Balls A and B roll across a table, then collide and bounce off each other. The paths of the two balls are pictured (viewed from above) in the diagram. (Figure 1) a) Which set of arrows best represents the change in momentum for balls A and B?b) Which of the following arrows indicates the direction of the impulse applied to ball A by ball B?

You know that a collision must be "elastic" if:a) The colliding objects stick togetherb) The sum of the final kinetic energies equals the sum of the initial kinetic energiesc) There is no change in the internal energies of the objects (thermal energy, vibrational energy, etc.)d) The colliding objects are stretchy or squishye) The momentum of the two-object system doesn't changeWhich one is true/false

Two billiard balls with the same mass m move in the direction of one another. Ball one travels in the positive x-direction with a speed of V1i, and ball two travels in the negative x-direction with a speed of V2i. The two balls collide elastically, and both balls change direction after collision. If the initial speeds of the balls were v1i=2.0m/s and v2i =1.0m/s, what would be the final speed and direction of ball two, v2f, in m/s?

For an elastic collision, which of the following statements are true? Choose all that apply.a) Kinetic energy is conserved.b) Momentum is gained.c) Kinetic energy is gained.d) Momentum is lost.e) Momentum is conserved.f) Kinetic energy is lost.

A baseball of mass m1 = 0.39 kg is thrown at another ball hanging from the ceiling by a length of string L = 1.95 m. The second ball m2 = 0.85 kg is initially at rest while the baseball has an initial horizontal velocity of V1 = 3.5 m/s. After the collision, the first baseball falls straight down (no horizontal velocity.)What is the angle that the string makes with the vertical at the highest point of travel in degrees?

As shown in the figure (Figure 1) , a superball with mass m equal to 50 grams is dropped from a height of hi =1.5 m. It collides with a table, then bounces up to a height of hf = 1.0 m. The duration of the collision (the time during which the superball is in contact with the table) is tc = 15ms. In this problem, take the positive y direction to be upward, and use g = 9.8 m/s2 for the magnitude of the acceleration due to gravity. Neglect air resistance.a. Find the y component of the momentum, pbefore,y, of the ball immediately before the collision.b. Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table.c. Find Jy, the y component of the impulse imparted to the ball during the collision.d. Find the y component of the time-averaged force Favg,y, in newtons, that the table exerts on the ball.e. Find Kafter - Kbefore, the change in the kinetic energy of the ball during the collision, in joules.

Block 1, of mass m1, moves across a frictionless surface with speed ui. It collides elastically with block 2, of mass m2, which is at rest (vi = 0). (Figure 1) After the collision, block 1 moves with speed uf, while block 2 moves with speed vf. Assume that m1 > m2, so that after the collision, the two objects move off in the direction of the first object before the collision.Part AWhat is the final speed uf of block 1? Express uf in terms of m1, m2, and ui.uf = ?Part BWhat is the final speed vf of block 2? Express vf in terms of m1, m2, and ui.vf = ?

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