Conservation of momentum and kinetic energy

In summary, two 0.3 Kg gliders collide elastically on a frictionless track, with a total kinetic energy of 0.52 J and a total momentum of 0.12 Kg m/s [left] along the track. By using the equations for conservation of momentum and energy, the final velocities of the gliders can be calculated. After subbing into the equations and solving for the final velocities, the resulting velocities should be 1.5 m/s [left] and 1.1 m/s [right]. It is important to note that when plugging in the relative velocities, √a²+b² is not the same as a+b, and when solving for vf1, the formula needs to be
  • #1
cherioslover
4
0

Homework Statement



Two 0.3 Kg gliders collide elastically on a frictionless track. Prior to the collision , their total kinetic energy is 0.52 J and their total momentum is 0.12 Kg m/s
along the track. Calculate the final velocities of the gliders.

Homework Equations


m1vi1 + m2vi2 = m1vf1 + m2vf2
1/2m1vi1^2 + 1/2m2vi2^2 = 1/2m1vf1^2 + 1/2m2vf2^2



The Attempt at a Solution



tried subbing into conservation of momentum equation, and the conservation of energy formula, then subbing 1 into 2 to find final velocities, but its not working. Can someone tell me what I am doing wrong or if the solution i am using is correct?

m1vi1 + m2vi2 = m1vf1 + m2vf2
-0.12 = (0.3)vf1 + (0.3)vf2
-0.4 = vf1 + vf2
-0.4 - vf2 = vf1

1/2m1vi1^2 + 1/2m2vi2^2 = 1/2m1vf1^2 + 1/2m2vf2^2
0.52 = (0.3)(vf1)^2 + (0.3)(vf2)^2
√(0.52/0.3) = vf1 + vf2
1.32= vf1 + vf2

sub 1 into 2

1.32 = -0.4 - vf2 + vf2

This is where i am confused, because vf2-vf2 is obviously 0. I know somethings not right, and some assistance would be kindly appreciated :) The answers if needed are supposed to be 1.5 m/s
and 1.1 m/s
 
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  • #2
line 3 in your KE paragraph: √a²+b² is NOT a+b .
you might square vf1 's formula , to plug into the KE line 2 ... eliminating it ... then solve for vf2.

otherwise, try subtracting the center-of-mass velocity, to get relative velocities. the relative velocity after collision is negative of the relative velocity before collision (since elastic; otherwise it is -e times vrel , with 0<e<1).
 

1. What is conservation of momentum?

Conservation of momentum is a fundamental law of physics that states that the total momentum of a closed system remains constant over time. This means that the total amount of movement or motion in a system will not change unless an external force is applied.

2. How is momentum conserved?

Momentum is conserved through interactions between objects. When two objects collide, the total momentum of the system before and after the collision remains the same. This is because, in a closed system, the forces between the objects are equal and opposite, resulting in a transfer of momentum from one object to the other.

3. What is kinetic energy?

Kinetic energy is the energy an object has due to its motion. It is directly proportional to an object's mass and the square of its velocity. The faster an object is moving, the more kinetic energy it possesses.

4. Is kinetic energy conserved?

No, kinetic energy is not conserved in a closed system. In collisions, some of the kinetic energy is converted into other forms of energy, such as heat or sound. However, the total energy in a closed system, including kinetic energy, remains constant.

5. How are momentum and kinetic energy related?

Momentum and kinetic energy are related through the equation E = 1/2mv^2, where E is kinetic energy, m is mass, and v is velocity. This equation shows that the kinetic energy of an object increases with its mass and velocity, both of which also contribute to its momentum.

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