How Do Elastic Collisions Affect Velocities and Center of Mass Movement?

In summary, in an elastic collision between a 0.206 kg ball with a velocity of 1.54 m/s and a 0.291 kg ball with a velocity of -0.396 m/s, the final velocities are 1.127 m/s and -0.263 m/s respectively. The velocity of the center of mass before the collision is 0.308 m/s and after the collision is 0.308 m/s. The conservation of momentum and energy equations were used to solve for the final velocities.
  • #1
DatAshe
9
0

Homework Statement


A ball of mass 0.206 kg with a velocity of 1.54 m/s meets a ball of mass 0.291 kg with a velocity of -0.396 m/s in a head-on, elastic collision.
(a) Find their velocities after the collision.
1f = m/s
2f = m/s
(b) Find the velocity of their center of mass before and after the collision.
cm, before = m/s
cm, after = m/s



Homework Equations


conservation of momentum? p=mv, pf=pi+I, KEf=KEi (because its elastic), Vcm=Ʃmivi/Ʃmi


The Attempt at a Solution


I have spent hours fiddling around with equations and I can't get anything to result in a correct answer. I am stumped and frustrated. This is past homework and I am just looking for an explanation on how I get from this information to the final answers. Step by step would be much appreciated, especially if you use calculus because I am lost and I am only to derivatives in calc I so far. Thank you for any and all help!
 
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  • #2
Yes, conservation of energy and of momentum.
What is the total momentum before? Total energy before?
In terms of the unknown final velocities, what are the energy and momentum after collision?
What equations do you get?
 
  • #3
wouldn't the before and after be the same do to conservation of momentum and energy?
so mivi^2=mfvf^2 and mivi=mfvm?
 
  • #4
You have to calculate the total momentum and energy.

That is the momentum of both balls added together. The sum total is the same before and after the collision. m1*v1i + m2*v2i = m1*v1f + m2*v2f

Same with energy.

That gives you a couple of equations with two unknowns (v1f and v2f) to solve.
 
Last edited:
  • #5


I would approach this problem by first defining the scenario and the variables involved. In this case, we have two balls of different masses colliding in a head-on, elastic collision. The masses of the balls are 0.206 kg and 0.291 kg, and their initial velocities are 1.54 m/s and -0.396 m/s respectively. We are asked to find the final velocities of the balls and the velocity of their center of mass before and after the collision.

Next, I would use the principles of conservation of momentum and conservation of kinetic energy to solve this problem. These principles state that in an isolated system (where there are no external forces acting), the total momentum and total kinetic energy remain constant before and after a collision.

First, we can calculate the total momentum before the collision. We can do this by multiplying the mass of each ball by its initial velocity and adding them together. In this case, we have:

initial momentum = (0.206 kg)(1.54 m/s) + (0.291 kg)(-0.396 m/s) = 0.318 kgm/s

Since this is an elastic collision, the total momentum after the collision will also be equal to 0.318 kgm/s. Now, we can use this fact to solve for the final velocities of the balls.

We can define the final velocities as v1f and v2f for the first and second ball respectively. Using the conservation of momentum principle, we can set up the following equation:

m1v1i + m2v2i = m1v1f + m2v2f

Substituting the values we know, we get:

(0.206 kg)(1.54 m/s) + (0.291 kg)(-0.396 m/s) = (0.206 kg)v1f + (0.291 kg)v2f

Solving for v1f and v2f, we get:

v1f = (0.206 kg)(1.54 m/s) + (0.291 kg)(-0.396 m/s) - (0.291 kg)(0.206 kg)(-0.396 m/s) = 0.872 m/s

v2f = (0.206 kg)(1.54 m/s) + (0.291 kg)(-0.396 m/s) - (0.206
 

1. How is the momentum conserved in an elastic ball collision?

In an elastic ball collision, the total momentum of the two balls before the collision is equal to the total momentum of the two balls after the collision. This means that the balls will have the same combined velocity after the collision as they did before the collision. This is known as the law of conservation of momentum.

2. What factors affect the elasticity of a ball?

The elasticity of a ball is affected by its material, shape, and temperature. Materials that are more flexible, such as rubber, will have a higher elasticity than materials that are less flexible. The shape of the ball also plays a role, as a rounder shape allows for more elastic deformation. Additionally, temperature can affect the elasticity of a ball, as colder temperatures can make materials more brittle and less elastic.

3. How is kinetic energy conserved in an elastic ball collision?

In an elastic ball collision, the total kinetic energy of the two balls before the collision is equal to the total kinetic energy of the two balls after the collision. This means that the energy is transferred between the two balls without any loss or gain. This is known as the law of conservation of energy.

4. What is the difference between an elastic and inelastic ball collision?

In an elastic ball collision, the two balls bounce off each other with no loss of energy. This means that the balls have the same kinetic energy before and after the collision. In an inelastic ball collision, some energy is lost as heat or sound, and the balls do not bounce back with the same velocity as before the collision.

5. How are the angles of incidence and reflection related in an elastic ball collision?

In an elastic ball collision, the angle of incidence (the angle at which the first ball hits the second ball) is equal to the angle of reflection (the angle at which the second ball bounces off the first ball). This is known as the law of reflection and is a result of the conservation of momentum and energy in the collision.

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