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Collisions -- conceptual questions

  • #1
2
0

Homework Statement


Let two particles of equal mass m collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary.

A: If the collision is elastic, what are the final velocities v_1 and v_2 of particles 1 and 2?

B: Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities v_1 and v_2 of particles 1 and 2?

Homework Equations


i:[/B] mv_1 + mv_2 = mv
ii: 0.5m(v_1)^2 + 0.5m(v_2)^2 = 0.5mv^2
iii: 2mv_1 + mv_2 = 2mv
iv: 0.5(2m)(v_1)^2 + 0.5m(v_2)^2 = 0.5(2m)v^2

The Attempt at a Solution


For A, after factoring out m and rearranging equation i to solve for v_2 I replaced v_2 in equation i with v_2 = (v - v_1) to yield:
v_1 + v_2 = v_1 + (v - v_1) = v_1 - v_1 + v = v, therefore
v = v.

I guessed (correctly) that v_1 = 0 and v_2 = v, but I'm not sure how I can find those answers using this equation.

It's the same thing with part B.

I tried rearranging the kinetic energy equations ii and iv for both parts A and B but still came out with such useless results as v^2 = v^2.

Why does particle 1 in part A transfer all of its momentum to particle 2?
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
15,494
1,876

Homework Statement


Let two particles of equal mass m collide. Particle 1 has initial velocity v, directed to the right, and particle 2 is initially stationary.

A: If the collision is elastic, what are the final velocities v_1 and v_2 of particles 1 and 2?

B: Now assume that the mass of particle 1 is 2m, while the mass of particle 2 remains m. If the collision is elastic, what are the final velocities v_1 and v_2 of particles 1 and 2?

Homework Equations


i:[/B] mv_1 + mv_2 = mv
ii: 0.5m(v_1)^2 + 0.5m(v_2)^2 = 0.5mv^2
iii: 2mv_1 + mv_2 = 2mv
iv: 0.5(2m)(v_1)^2 + 0.5m(v_2)^2 = 0.5(2m)v^2

The Attempt at a Solution


For A, after factoring out m and rearranging equation i to solve for v_2 I replaced v_2 in equation i with v_2 = (v - v_1) to yield:
v_1 + v_2 = v_1 + (v - v_1) = v_1 - v_1 + v = v, therefore
v = v.

I guessed (correctly) that v_1 = 0 and v_2 = v, but I'm not sure how I can find those answers using this equation.
Substitute v_2 = (v - v_1) into the energy equation.

It's the same thing with part B.
Again, use both equations, momentum and energy. Express v_2 with v_1 from the momentum equation and substitute into the energy equation.
I tried rearranging the kinetic energy equations ii and iv for both parts A and B but still came out with such useless results as v^2 = v^2.

Why does particle 1 in part A transfer all of its momentum to particle 2?
It is the solution of the system of equation i and ii.
 
  • #3
2
0
Substitute v_2 = (v - v_1) into the energy equation.


Again, use both equations, momentum and energy. Express v_2 with v_1 from the momentum equation and substitute into the energy equation.

It is the solution of the system of equation i and ii.
That makes sense, thanks!
 

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