What was the mistake in solving for the speeds of two marbles after a collision?

  • #1
anotherphysicsgeek
3
1
Homework Statement
A 49.0 g marble moving at 1.90 m/s strikes a 28.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.

What is the speed of 49.0 g marble immediately after the collision?
What is the speed of 28.0 g marble immediately after the collision?
Relevant Equations
1. m1v1i+m2v2i=m1v1f+m2v2f
2. 1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2
Solved equation 1 for v1f and then substituted into equation 2 and solved for v2f. Got 2.22 as the answer, but it said the answer is incorrect.
 
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  • #2
anotherphysicsgeek said:
Homework Statement: A 49.0 g marble moving at 1.90 m/s strikes a 28.0 g marble at rest. Note that the collision is elastic and that it is a "head-on" collision so all motion is along a line.

What is the speed of 49.0 g marble immediately after the collision?
What is the speed of 28.0 g marble immediately after the collision?
Relevant Equations: 1. m1v1i+m2v2i=m1v1f+m2v2f
2. 1/2m1v1i^2+1/2m2v2i^2=1/2m1v1f^2+1/2m2v2f^2

Got 4.4333 as the answer
This is the answer to what? There are two questions.
 
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  • #3
Hill said:
This is the answer to what? There are two questions.
Sorry it was the answer to the second question. Also going back and re-plugging in the numbers I got 2.22, but it's still wrong.
 
  • #4
anotherphysicsgeek said:
Sorry it was the answer to the second question. Also going back and re-plugging in the numbers I got 2.22, but it's still wrong.
You need to show your work. Also, take a look at the Latex guide for maths equations, under the help pages.

In a question like this, where one mass is initially at rest, I would automatically simplify the relevant Equations. If you reply to this post, you'll see the Latex I used. I prefer ##u, v## to ##v_i, v_f##. E.g.
$$m_1u_1 = m_1v_1+m_2v_2$$$$\frac 1 2 m_1u_1^2 = \frac 1 2 m_1v_1^2 +\frac 1 2 m_2v_2^2$$
 
  • #5
anotherphysicsgeek said:
Sorry it was the answer to the second question. Also going back and re-plugging in the numbers I got 2.22, but it's still wrong.
That's not far off. Maybe a rounding error. Keep at least three sig figs until the end.
But it is silly having to guess. As @PeroK says, post your working.
 
  • #6
Thank you guys, and I'll fix that in the future. I ended up finding the correct answer and it was an algebra mistake.
 
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1. What is the equation used to calculate the speeds of two marbles after a collision?

The equation used is the conservation of momentum equation, which states that the total momentum before the collision is equal to the total momentum after the collision.

2. What is the most common mistake made when solving for the speeds of two marbles after a collision?

The most common mistake is not taking into account the direction of the velocities. This can result in incorrect calculations and an inaccurate understanding of the collision.

3. How can one ensure that the correct values are used when solving for the speeds of two marbles after a collision?

To ensure accuracy, it is important to label the velocities and masses of the marbles correctly and to use the correct signs (+/-) in the momentum equation. It is also helpful to draw a diagram to visualize the collision and the direction of the velocities.

4. Can the speeds of the marbles be calculated if the masses are not known?

No, the masses of the marbles are necessary in order to calculate their speeds after the collision. Without this information, the conservation of momentum equation cannot be solved.

5. Is it possible to have a perfectly elastic collision between two marbles?

Yes, a perfectly elastic collision is one in which there is no loss of kinetic energy. This can occur if the marbles have the same mass and the collision is head-on, with no friction or external forces acting on the system.

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