Solving Collision Problem: 10g & 15g Objects Moving in Opposite Directions

  • Thread starter the7joker7
  • Start date
  • Tags
    Collision
In summary: No, you did not make a math error.You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.
  • #1
the7joker7
113
0

Homework Statement



1) A 10.0-g object moving to the right at 20 cm/s makes an elastic head-on collision with a 15.0-g object moving in the opposite direction at 30.0 cm/s. Find the velocity of each object after collision.

Homework Equations



m[tex]_{1}[/tex]v[tex]_{i1}[/tex] + m[tex]_{2}[/tex]v[tex]_{i2}[/tex] = m[tex]_{1}[/tex]v[tex]_{f1}[/tex] + m[tex]_{2}[/tex]v[tex]_{f2}[/tex]

(.5)m[tex]_{1}[/tex]v[tex]_{1i}[/tex][tex]^{2}[/tex] + .(5)m[tex]_{2}[/tex]v[tex]_{2i}[/tex][tex]^{2}[/tex] = (.5)m[tex]_{1}[/tex]v[tex]_{1f}[/tex][tex]^{2}[/tex] + .(5)m[tex]_{2}[/tex]v[tex]_{2f}[/tex][tex]^{2}[/tex]

The Attempt at a Solution



I plugged values into the equations and got...

(.01)(.2) + (.015)(-.3) = (.01)(v[tex]_{f1}[/tex]) + (.015)(v[tex]_{f2}[/tex])

(.5)(.01)(.2^2) + (.5)(.015)(-.3^2) = (.5)(.01)(v[tex]_{f1}[/tex][tex]^{2}[/tex]) + (.5)(.015)(v[tex]_{f2}[/tex][tex]^{2}[/tex])

But I'm still stuck with two variables. How do I solve one of them?
 
Physics news on Phys.org
  • #2
What is conserved in an elastic collision? (momentum and...)
 
  • #3
hunter151 said:
What is conserved in an elastic collision? (momentum and...)

Energy.

I was under the impression I had the equation for conservation of energy there (the 2nd one).
 
  • #4
Oh wow, I am an idiot. You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.

For example, you could find Vf1 in terms of Vf2 from the first equation, and plug this Vf1 into the second equation. Now you are faced with an equation of just Vf2 as unknown.
 
  • #5
hunter151 said:
Oh wow, I am an idiot. You are correct on the equations. Now you're faced with two equations and two unknowns; therefore, you can solve. One way of doing things is to solve one of the equations for one variable, then plug this result into the second equation, to be left with a single equation with one unknown variable.

For example, you could find Vf1 in terms of Vf2 from the first equation, and plug this Vf1 into the second equation. Now you are faced with an equation of just Vf2 as unknown.

Thanks.

I'm trying to do that now, and...

.0002 - .000675 = (.005)vf1^2 + (.0075)vf2^2

.005vf1^2 = -.000475 - .0075vf2^2

vf1^2 = .095 - 1.5vf2^2

vf1 = .3082 - 1.5vf2

-.0025 = (.01)(.3082 - 1.5vf2) + (.015)(vf2)

-.005582 = -(.015vf2) + (.015vf2)

Well, my vf2s cancel out :/

Did I make a math error?
 
  • #6
When you took the square root of Vf1, you forgot that there are two solutions, + and -. ;)
 

Related to Solving Collision Problem: 10g & 15g Objects Moving in Opposite Directions

1. What is a collision problem?

A collision problem is when two or more objects come into contact with each other and transfer energy. This can occur when objects are moving in the same or opposite directions.

2. How do you solve a collision problem?

To solve a collision problem, you must first identify the initial velocities and masses of the objects involved. Then, you can use the principles of conservation of momentum and conservation of energy to calculate the final velocities and any changes in kinetic energy.

3. What is conservation of momentum?

Conservation of momentum states that the total momentum of a closed system remains constant, meaning the total momentum before a collision is equal to the total momentum after the collision. This principle is used to solve collision problems involving objects with different masses and velocities.

4. What is conservation of energy?

Conservation of energy states that energy cannot be created or destroyed, only transferred from one form to another. In the case of a collision problem, this means that the total kinetic energy before the collision is equal to the total kinetic energy after the collision.

5. How do you determine the direction of the final velocities in a collision problem?

The direction of the final velocities in a collision problem can be determined by using the principle of conservation of momentum. The final velocities will depend on the initial velocities and masses of the objects, as well as the type of collision (elastic or inelastic).

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
5K
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
2
Views
13K
Replies
5
Views
5K
  • Introductory Physics Homework Help
Replies
3
Views
1K
Back
Top