of which i thought was an ellastic collision problem

[rottenapple]

Homework Statement

Ball A moving at 12m/s collides elastically with ball B as shown. If both balls have the same mass, what is the final velocity of ball A? ***theta = 60***

Homework Equations

Ui + Ki = Uf + Kf

The Attempt at a Solution

I could only think of the above equation when I read elastic collision. However, the solution in the back of the book suggested using conservation of momentum, which I thought is only applicable to inelastic collision. The numerical answer is 6m/s which can be figure out using mv = mv and breaking it down to x and y components.
I can see how they get to the numerical answer but I am lost as to why? Can someone please explain why I would use the formula for conservation of momentum? And also, I'm at a complete loss on how to approach a problem like this so can you also explain what I would need to solve the problem if it is an elastic collision?

Thank you!

 

[diazona]

Conservation of momentum applies to all collisions. Conservation of energy applies to only elastic collisions.

So in summary,
Elastic:
$$K_i + U_i = K_f + U_f$$
$$\vec{p}_i = \vec{p}_f$$

Inelastic:
$$\vec{p}_i = \vec{p}_f$$ only

 

[ben.zhang98]

The forces due to the collision among particles(interior forces) are always larger than exterior forces, and they are so large that we can always omit the exterior forces. According to Newton's second law, F=d(mv)/dt, if F(exterior force) is zero , mv(momentum) must be a constant, i.e. momentum is conservertive. So the conservertion of momentum on collision is always valid.

 

[inky]

Homework Statement

Ball A moving at 12m/s collides elastically with ball B as shown. If both balls have the same mass, what is the final velocity of ball A? ***theta = 60***

Homework Equations

Ui + Ki = Uf + Kf

The Attempt at a Solution

I could only think of the above equation when I read elastic collision. However, the solution in the back of the book suggested using conservation of momentum, which I thought is only applicable to inelastic collision. The numerical answer is 6m/s which can be figure out using mv = mv and breaking it down to x and y components.
I can see how they get to the numerical answer but I am lost as to why? Can someone please explain why I would use the formula for conservation of momentum? And also, I'm at a complete loss on how to approach a problem like this so can you also explain what I would need to solve the problem if it is an elastic collision?

Thank you!

For elastic collision, you may use both law of conservation of momentum and law of conservation of energy.
Ui + Ki = Uf + Kf

For inelastion collision , you can use only law of conservation of momentum. In inelastic collision , energy lost or gained. If you want to use law of conservation of energy,
Ui + Ki= Uf + Kf +energy lost

 

[rdl]

I would like to explain the reasoning behind using conservation of momentum in this problem. The conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system (where no external forces act upon the system) remains constant. In an elastic collision, the total kinetic energy of the system is conserved, which means that the total momentum of the system must also be conserved.

In this problem, we have two balls colliding with each other, and no external forces are acting on the system. Therefore, we can use the conservation of momentum to solve for the final velocity of ball A. We know that both balls have the same mass, so we can simplify the equation to:

m*vA + m*vB = m*vA' + m*vB'

Where vA and vB are the initial velocities of balls A and B, and vA' and vB' are their final velocities.

Since the problem provides us with the value of theta, we can use trigonometry to break down the velocities into their x and y components. This will give us two equations:

m*vA*cos(60) + m*vB*cos(0) = m*vA'*cos(30) + m*vB'*cos(90)

m*vA*sin(60) + m*vB*sin(0) = m*vA'*sin(30) + m*vB'*sin(90)

Simplifying these equations, we get:

0.5*m*vA + m*vB = 0.5*m*vA' + m*vB'

0.87*m*vA = 0.5*m*vA' + m*vB'

We can now solve for vA' by substituting the value of vB' from the first equation into the second equation:

0.87*m*vA = 0.5*m*vA' + 0.5*m*vA

Solving for vA', we get:

vA' = 0.87*vA

Since vA = 12m/s, the final velocity of ball A is vA' = 0.87*12 = 10.44m/s, which is approximately 10.4m/s.

In summary, we used the conservation of momentum to solve for the final velocity of ball A in this elastic collision problem. It is important to understand the fundamental principles behind the equations we use in physics