Head-on elastic collision between two masses

In summary, the experimenter wants the incident mass to lose a quarter of its kinetic energy in the elastic collision. To find the ratio of masses, you can assume one mass is 1 and the other is x. Use conservation of energy and momentum to find the speeds after the collision, which will depend on both x and the initial speed v0. Then, solve for x so that the moving mass loses a quarter of its kinetic energy.
  • #1
mangafan
4
0
in a head-on elastic collision between two masses, one of which is stationary, a experimenter would like the incident mass to lose quarter of its kinetic energy in the collision. What must be the ratio of the masses of the two objects for this to be the case?
 
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  • #2
First you need to ask yourself: what are the conditions of elastic collisions? there are some keys that will help you write some mathematical relationships.

Note: You need to show some work to us before we'll help. If you're stuck, you should look at related threads.
 
  • #3
Good question! What have you done so far?

Since you are only interested in the ratio of masses, you can assume the mass of one is 1 and the other x. Then x is the "ratio of masses" you want. Now, one of them (it doesn't matter which) is stationary (v= 0) and the other has some speed v0. What is the total kinetic energy? Use conservation of energy and conservation of momentum to find the speeds after the collision (which will, of course, depend on both x and v0) so you can find the kinetic energy of each. What should x be so that one that was moving has lost 1/4 of its kinetic energy?
 
  • #4
I have use the equation v1=(m1-m2)*u1/(m1+m2) and v2=(2m1)*u1/(m1+m2)
then i substitute v1 and v2 into the ke equation. i also substitute m1 = 1 and m2 = x. i did not get the result i want
 
  • #5
i use two equation v1= (m1-m2)*u1/(m1+m2) and v2= (2m1)*u1/(m1+m2). i substitute m1 = 1 and m2 = x. and then substitute in the ke equation. but i just cannot get the value for x
 

1. What is a head-on elastic collision between two masses?

A head-on elastic collision between two masses is a type of collision where two objects collide directly into each other, conserving both momentum and kinetic energy. Elastic collisions are characterized by the objects bouncing off of each other without any loss of energy.

2. How is momentum conserved in a head-on elastic collision?

Momentum is conserved in a head-on elastic collision through the principle of conservation of momentum. This means that the total momentum of the two objects before the collision is equal to the total momentum after the collision. This holds true for all types of collisions, including head-on elastic collisions.

3. Can you calculate the final velocities of the two masses in a head-on elastic collision?

Yes, the final velocities of the two masses in a head-on elastic collision can be calculated using the equations of conservation of momentum and conservation of kinetic energy. These equations take into account the masses and initial velocities of the objects, as well as the coefficient of restitution, which represents the elasticity of the collision.

4. What is the coefficient of restitution and how does it affect a head-on elastic collision?

The coefficient of restitution is a measure of the elasticity of a collision. It is a ratio of the final relative velocity of two objects after a collision to the initial relative velocity before the collision. In a head-on elastic collision, the coefficient of restitution is equal to 1, meaning that the objects bounce off of each other with the same relative speed as they approached each other.

5. Are there any real-life examples of head-on elastic collisions?

Yes, there are many real-life examples of head-on elastic collisions. One example is a game of billiards, where the balls collide with each other and bounce off with the same relative speed. Another example is a trampoline, where a person jumps and bounces off the surface. In both cases, the collisions are nearly elastic, with minimal loss of energy.

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