Momentum question involving an elastic collision

In summary, when two skaters collide, their momentum and kinetic energy are conserved, but their elastically-generated energy is not.
  • #1
Heron
3
0
1. A 62 kg skater traveling at 2.8 m/s accidentally bumps elastically into a 76 kg skater moving at 1.6m/s in the same direction
Find the velocity of each skater after the collision




2. m1v1 + m1v2 = m1v3 + m2v4



3.
 
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  • #2
As you stated, momentum is conserved but so is one other thing. Can you figure out what?
 
  • #3
Kinetic energy is also conserved.

This is all the work I have done so far.

(62)(2.8) + (76)(1.6) = 62v1 + 76v2

0.5(62)(2.8)^2 + 0.5(76)(1.6)^2 = 0.5(62)(v1)^2 + 0.5(76)(v2)^2

295.2 = 62v1 + 76v2 and 340.32 = (31)(v1)^2 + (38)(v2)^2

v2= (295.2 - 62v1)/76

340.32 = (31)(v1)^2 + (38)[(295.2 - 62v1)/76]^2

0 = (31)(v1)^2 + 357.778v1 + (25.289)(v1)^2 + 573.30947 - 340.32
 
Last edited:
  • #4
It all looks good except the last line. In fact, just the term with v1 looks incorrect. Try checking the bracket expansion again?
 
  • #5
I fixed it.
When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.
 
  • #6
Heron said:
I fixed it.
When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.
Did you substitute these back to verify that your answers correctly solve your equation?

There is always the option of using an online calculator to check your answer, to give yourself confidence that you have the solution, e.g., http://www.quickmath.com/webMathema...+(31)(v1)^2+++(38)((295.2+-+62v1)/76)^2&v2=v1

Note: Mathematica ignores square brackets, so you must change them to round parentheses ().
 
  • #7
If you set up the equation correctly, one root has to be the same as the original velocity.

You can follow a much simpler method of solving.

Arrange the equations both for momentum and energy so that the initial and final velocity of one body is at one side and those of the other body on the other side:

m1(u1-v1)=m2(v2-u2)
m1(u12-v12)=m2(v22-u22).


As v≠u for either body, you can divide the second equation with the first one, and obtain a simple first-order system of equations

u1+v1=u2+v2,
m1(u1-v1)=m2(v2-u2).


ehild
 

1. What is momentum in an elastic collision?

Momentum is a measure of an object's motion and is defined as the product of its mass and velocity. In an elastic collision, the total momentum of the system is conserved, meaning that the total momentum before the collision is equal to the total momentum after the collision.

2. How is momentum conserved in an elastic collision?

In an elastic collision, momentum is conserved through the exchange of kinetic energy between two objects. When two objects collide, the total momentum of the system remains the same before and after the collision. This means that the sum of the momentum of each individual object before the collision is equal to the sum of their momentum after the collision.

3. What is the formula for calculating momentum in an elastic collision?

The formula for calculating momentum in an elastic collision is p = mv, where p is momentum, m is mass, and v is velocity. This formula can be applied to both objects involved in the collision, and their individual momenta can be added together to find the total momentum of the system.

4. Can momentum be transferred between objects in an elastic collision?

Yes, momentum can be transferred between objects in an elastic collision. When two objects collide, they exert forces on each other, causing a transfer of momentum. This transfer of momentum results in changes in the objects' velocities.

5. What happens to the kinetic energy in an elastic collision?

In an elastic collision, kinetic energy is conserved. This means that the total kinetic energy of the system before the collision is equal to the total kinetic energy after the collision. However, the kinetic energy may be distributed differently between the two objects involved in the collision.

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