# Inelastic collision Two angles and final velocities

• -PhysicsMajor-
In summary: This is correct. Now, using the fact that ##V_1## and ##V_2## are related by the condition that the final kinetic energy is ##(1-Q)KE_{initial}##, you should be able to solve for ##V_1## and ##V_2## in terms of ##\theta##, ##\phi##, and ##Q##. Then, you can use this to find ##\phi## by realizing that the change in x-velocity for ##M_2## is related to ##\phi##.In summary, for an inelastic collision between two blocks on a horizontal plane, the final kinetic energy can be found using the equation KE final = (1-Q)KE initial.
-PhysicsMajor-

## Homework Statement

Consider an inelastic collision between two blocks on a horizontal plane. Block M1 is moving with velocity Vo and collides with block M2 which is at rest. During the collision a fraction Q of the original kinetic energy is lost. It is observed that M1 is deflected by an angle theta (above the x axis), and M2 is deflected at an angle phi (below the x axis). After the collision M1 is moving to the right.

Find the angle phi, and the final velocities of the blocks.

M1=2kg
M2=4kg
Vo=10m/sec
Theta=30o
Q=.2

## Homework Equations

Inital Momentum = Final Momentum (M1iVo+M2iVo =M1fVf+M2fVf)

KEf = (1-Q)KEi

*I'm sure there are more but this is what I have from class.

## The Attempt at a Solution

Finding the final KE was easy enough using the above equation: KE inital =100 KE final =80

Then I tried to split them up in the x and y components. They both start with zero in the y direction and only M1 has initial momentum in the x direction. In the Y direction after the impact I got some sin and cos directions
So M1iVo = M1V1fCos(theta)+M2V2fcos(phi)+M1V1f-M2V2fsin(phi) (negative because it is below the x axis)
If that is right (big if) I have no clue where to go from here. I really don't know how to turn the final kinetic energy into two different objects and I REALLY don't know how to find that phi angle. I'm at a complete loss.

This is my first post here so hopefully formatted the question correctly. This stuff is really stressing me out. Every time I think that I've stumbled upon something useful online, it just seems way to simple, which is pretty much a guarantee that it can't help me. Our homework consists of just a few problems over the course of a week, and when he shows examples in class my professor fills several chalkboards full of calculations. So anything that is a quick answer must be wrong.

I don't want the answer, I just need a nudge in the right direction of how to even set up the problem. I have not found a single thing online that can help with these intermediate inelastic collisions with two objects of a different mass.

This forum is my last hope.

*Way too many words, not nearly enough physics. Sorry

#### Attachments

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-PhysicsMajor- said:
So M1iVo = M1V1fCos(theta)+M2V2fcos(phi)+M1V1f-M2V2fsin(phi)
you seem to have a basic misapprehension regarding momentum.
Momentum is a vector, and your 'relevant equation' for its conservation should be interpreted in that light.
When resolving into separate X and Y directions, that gives you two momentum conservation equations, one for each direction. But you appear to have collapsed them into one scalar equation.

haruspex said:
you seem to have a basic misapprehension regarding momentum.
Momentum is a vector, and your 'relevant equation' for its conservation should be interpreted in that light.
When resolving into separate X and Y directions, that gives you two momentum conservation equations, one for each direction. But you appear to have collapsed them into one scalar equation.
Thank you for bringing this up, because it seemed wrong when I was doing it. I guess I just got confused because there is no inital momentum in the y direction.

So
Pix=Pfx = M1iVo = M1V1fCos(theta)+M2V2fcos(phi)
Piy = Pfy = 0 = M1V1fsin(theta)-M2V2fsin(phi)

??

-PhysicsMajor- said:
Thank you for bringing this up, because it seemed wrong when I was doing it. I guess I just got confused because there is no inital momentum in the y direction.

So
Pix=Pfx = M1iVo = M1V1fCos(theta)+M2V2fcos(phi)
Piy = Pfy = 0 = M1V1fsin(theta)-M2V2fsin(phi)

??
Yes.

## What is an inelastic collision?

An inelastic collision is a type of collision in which kinetic energy is not conserved. This means that the total kinetic energy of the system before the collision is not equal to the total kinetic energy after the collision. In an inelastic collision, some of the kinetic energy is converted into other forms, such as heat or sound.

## What are the two angles involved in an inelastic collision?

The two angles involved in an inelastic collision are the angle of incidence and the angle of reflection. The angle of incidence is the angle at which the object approaches the other object before the collision, while the angle of reflection is the angle at which the object moves away from the other object after the collision.

## How are the final velocities in an inelastic collision calculated?

The final velocities in an inelastic collision can be calculated using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time taken for the collision to occur. However, the actual calculation of final velocities in an inelastic collision is more complex and involves taking into account the masses and velocities of both objects involved.

## What factors affect the final velocities in an inelastic collision?

The final velocities in an inelastic collision are affected by several factors, including the masses and velocities of the two objects involved, the angle of incidence, and the coefficient of restitution (a measure of the elasticity of the objects). In addition, external forces such as friction can also affect the final velocities.

## How does an inelastic collision differ from an elastic collision?

In an elastic collision, kinetic energy is conserved and the objects involved bounce off each other without any loss of energy. However, in an inelastic collision, kinetic energy is not conserved and some of the energy is converted into other forms. In addition, the objects involved in an inelastic collision may stick together after the collision, while in an elastic collision they will bounce off each other.

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