Conservation of Momentum and Energy

In summary, the conversation discusses a problem involving a head-on collision between a proton and a stationary carbon nucleus. The speed of the proton is given and the goal is to find its velocity after the collision. The participants mention that both momentum and energy are conserved in this scenario and provide equations to solve for the final velocity. However, the final calculated velocity of -678.4 m/s seems incorrect and one participant asks for any ideas on what they may be doing wrong.
  • #1
Lyuokdea
154
0
This one is giving me fits for some reason:

A proton of mass m undergoes a head-on collision with a stationary carbon nucleus of mass 12m. The speed of the proton is 300 m/s. Find the velocity of the proton after the collision.

Ok, I know that both momentum and energy are conserved. So:

mTvT = m1*v1 + m2v2

and

1/2mTvT^2 = 1/2 m1*v1^2 + 1/2 m2*v2^2

I tried to solve for v2 in both equations and then set them equal to each other:

v2 = (mTvT - m1v1) / m2

and
v2 = sqrt((1/2mTvT^2 - 1/2 m1v1^2)/m2)

and then solved for:
(mTvT - m1v1) / m2= sqrt((1/2mTvT^2 - 1/2 m1v1^2)/(.5*m2))

but this gives me -678.4 m/s which isn't right, apparently I'm doing something wrong here.

~Lyuokdea
 
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  • #2
any ideas?
 
  • #3



It seems like you have the right idea in terms of using the conservation of momentum and energy equations to solve for the final velocity of the proton after the collision. However, it looks like there may be a mistake in your calculation.

To begin with, the conservation of momentum equation should be written as:

m1v1 + m2v2 = (m1 + m2)v

where v is the final velocity of the two particles after the collision. In this case, m1 represents the mass of the proton and m2 represents the mass of the carbon nucleus. So the equation should look like this:

mv = (m + 12m)v2

Next, for the conservation of energy equation, you have the right idea in terms of setting the kinetic energy of the initial particle equal to the sum of the kinetic energies of the two particles after the collision. However, the equation should be written as:

1/2m1v1^2 + 1/2m2v2^2 = (1/2m1 + 1/2m2)v^2

Plugging in the values for this problem, the equation would look like this:

1/2mv^2 = (1/2m + 1/2*12m)(300)^2

Simplifying this, you should end up with:

v^2 = 300^2/7

Taking the square root of both sides, you should get the final velocity of the proton after the collision as:

v = 122.5 m/s

This is a positive value, indicating that the proton moves in the same direction as its initial velocity. It's possible that your mistake came from using the wrong values for m1 and m2 in your equations, so make sure to double check those as well. I hope this helps!
 

1. What is conservation of momentum and energy?

Conservation of momentum and energy is a fundamental law of physics that states that the total momentum and energy of a closed system remains constant over time. This means that the total amount of momentum and energy in a system cannot be created or destroyed, but can only be transferred or transformed.

2. How is conservation of momentum and energy related?

The conservation of momentum and energy are related in the sense that they are both fundamental properties of a system. Momentum is the product of an object's mass and velocity, while energy is the ability to do work. In a closed system, the total amount of momentum and energy will remain the same.

3. What are some examples of conservation of momentum and energy?

Some examples of conservation of momentum and energy include a pendulum swinging back and forth, a roller coaster going up and down, or a ball bouncing on the ground. In all of these scenarios, the total amount of momentum and energy remains constant, even though the forms of energy may change.

4. How does conservation of momentum and energy apply to real-world situations?

Conservation of momentum and energy applies to many real-world situations, such as collisions between objects, motion of planets in the solar system, and the flow of fluids in pipes. It is a fundamental principle that helps us understand and predict the behavior of physical systems in our everyday lives.

5. What are the implications of violating conservation of momentum and energy?

If conservation of momentum and energy is violated in a system, it means that there is an external force or energy input that is not accounted for. This can lead to unexpected and unpredictable behavior in the system. Violation of this law is rare, but can occur in extreme circumstances such as nuclear reactions or black holes.

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