Conservation of Momentum and Inelastic Collisions

[dominus96]

Homework Statement

Two identical objects, A and B, both with the same mass of M, move in the same direction (right) on a horizontal, frictionless surface. Object A is to the left of object B, and is moving at velocity 3V, while object B is moving at velocity V, so A will eventually collide with B.

a) Determine the total momentum of this system, in terms of M and V.

b) Assuming the collision is inelastic, what is the speed of the two objects after colliding?

Homework Equations

P=MV

The Attempt at a Solution

Since it's inelastic, the momentum and kinetic energy are conserved. But I don't know how to apply 2 masses since I'm only used to doing one.

 

[dominus96]

bump, anyone please help?

 

[hotcommodity]

For part a, simply write out the total momentum and plug M, V, and 3V into their respective components. The mass of each object is the same, so this helps to simpyify the equation. For part b, we have an inelastic collision, this means that the objects stick to each other, so kinetic energy is not conserved. However, you're told that friction does not act on the objects, and thus linear momentum will be conserved. Hint: After colliding, the two objects "stick" to each other, what does this say about their final velocity?

 

[dominus96]

Ok for part b since object A is moving faster at 3V, wouldn't the 2 objects just move also at 3V after they stick?

I still don't understand whayyou mean by part a. Do I need to write like MV+M3V? Because that would simplify to M4V, which doesn't seem right.

 

[PhanthomJay]

Homework Statement

Two identical objects, A and B, both with the same mass of M, move in the same direction (right) on a horizontal, frictionless surface. Object A is to the left of object B, and is moving at velocity 3V, while object B is moving at velocity V, so A will eventually collide with B.

a) Determine the total momentum of this system, in terms of M and V.

b) Assuming the collision is inelastic, what is the speed of the two objects after colliding?

Homework Equations

P=MV

The Attempt at a Solution

Since it's inelastic, the momentum and kinetic energy are conserved. But I don't know how to apply 2 masses since I'm only used to doing one.

No, that's not right. First off, the problem is not worded properly, so i can only assume that this is a totally inelastic collision, that is, the objects stick together when they collide. Momentum is always conserved during collisions, whether they are completely elastic, inelastic, or totally (completely) inelastic. Kinetic energy is conserved only in completely elastic collisions. All you need to know for this problem is the conservation of momentum equation.

 

[hotcommodity]

Ok for part b since object A is moving faster at 3V, wouldn't the 2 objects just move also at 3V after they stick?

Not at 3V no, but they would have the same final velocity. Can you show me the equation you would use to solve for the final velocity?

I still don't understand whayyou mean by part a. Do I need to write like MV+M3V? Because that would simplify to M4V, which doesn't seem right.

You've got it. Why doesn't this seem right to you?

 

[rl.bhat]

According to conservation of momentum M*3V + M*V = MV1 + MV2
Since collision is inelastic , coefficient of retitution e = (V2-V1)/(3V-V). Using these two equations find V1 and V2

 

[hotcommodity]

According to conservation of momentum M*3V + M*V = MV1 + MV2
Since collision is inelastic , coefficient of retitution e = (V2-V1)/(3V-V). Using these two equations find V1 and V2

V1 and V2 will be the same, and they can be found by the first equation in your post.

 

[dominus96]

Thank you guys I figured it out. There is a second part to the question that I could quickly use help on also:

a) Suppose the 2 objects perform a perfect elastic collision instead, so object A bounces back to the left. Object B now increases speed from V to 2.5V. What is the speed of object A after the collision?

b) What is the kinetic energy lost during this collision?

 

[hotcommodity]

Thank you guys I figured it out. There is a second part to the question that I could quickly use help on also:

a) Suppose the 2 objects perform a perfect elastic collision instead, so object A bounces back to the left. Object B now increases speed from V to 2.5V. What is the speed of object A after the collision?

b) What is the kinetic energy of this collision?

You'll want to use the first equation for conservation of momentum in rl.bhat's post. Now you know the final velocity of B, and only have one unknown, the final velocity of A.

Edit: To find the loss of kinetic energy, you'll want to find the initial kinetic energy of the system (A and B) and the final kinetic energy of the system, and take the difference between the two. The initial kinetic energy would have the form $$K_0 = .5(m_A + m_B)(v_{A0} + v_{B0})^2$$. But don't forget that you'll be dealing with a negative velocity after the collision.

 

[dominus96]

Ok thanks, what about part b?

 

[hotcommodity]

The hint on part b is in my edited post above. I have to get to bed, good luck! :)

 

[dominus96]

Ty very much.

 

[rl.bhat]

a) Suppose the 2 objects perform a perfect elastic collision instead, so object A bounces back to the left. Object B now increases speed from V to 2.5V. What is the speed of object A after the collision?

b) What is the kinetic energy lost during this collision?
In the perfect elastic collision, e = 1. Hence after collision VB - VA = 2V. Before collision we have VB + VA = 4V. Solving these two equations we get VA = V and VB = 3V. And there is no loss of KE.