Momentum question involving an elastic collision

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In an elastic collision between a 62 kg skater at 2.8 m/s and a 76 kg skater at 1.6 m/s, both momentum and kinetic energy are conserved. The equations for momentum and kinetic energy are set up to find the final velocities of each skater after the collision. Initial calculations led to incorrect values for the final velocity of the first skater, prompting a review of the algebra involved. Suggestions included verifying the substitution of values back into the equations and using online calculators for confirmation. A simpler method for solving the equations was also proposed, emphasizing the relationship between the initial and final velocities of the skaters.
Heron
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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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As you stated, momentum is conserved but so is one other thing. Can you figure out what?
 
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
 
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It all looks good except the last line. In fact, just the term with v1 looks incorrect. Try checking the bracket expansion again?
 
I fixed it.
When I solve it I get -0.7366 and -5.6195 for v1 but neither of those are answers.
 
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 ().
 
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
 

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