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

What is the common mistake when using the conservation of momentum?

A common mistake is not properly accounting for the direction of the velocities. Momentum is a vector quantity, so you need to consider both magnitude and direction. Failing to do so can lead to incorrect results.

How does incorrect mass assignment affect the outcome?

If the masses of the marbles are incorrectly assigned or swapped, the calculated velocities after the collision will be incorrect. Accurate mass measurements are crucial for applying the conservation of momentum and energy principles correctly.

What errors can occur when applying the conservation of kinetic energy?

One frequent error is assuming that kinetic energy is always conserved. This is only true for perfectly elastic collisions. In inelastic collisions, some energy is transformed into other forms, such as heat or sound, and not all kinetic energy is conserved.

Why is it important to correctly identify the type of collision?

Identifying whether the collision is elastic or inelastic is crucial because it determines whether kinetic energy is conserved. Misidentifying the collision type can lead to incorrect application of the conservation laws and thus incorrect results.

How does neglecting friction or other external forces lead to mistakes?

Neglecting external forces like friction can lead to inaccurate results because these forces can alter the momentum and energy of the system. Always consider whether external forces are significant enough to influence the outcome of the collision.

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