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

Click For Summary
The discussion centers on solving for the speeds of two marbles after an elastic collision. The initial calculations yielded an incorrect speed of 2.22 m/s for one marble, prompting confusion about which question it addressed. Participants emphasized the importance of showing work and suggested simplifying equations when one mass is at rest. A rounding error was noted as a potential issue, and ultimately, the correct answer was found to be due to an algebra mistake. Clear communication and detailed calculations are essential for solving such problems accurately.
anotherphysicsgeek
Messages
3
Reaction score
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.
 
Last edited:
Physics news on Phys.org
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.
 
Last edited:
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.
 
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.
$$m_1u_1 = m_1v_1+m_2v_2$$$$\frac 1 2 m_1u_1^2 = \frac 1 2 m_1v_1^2 +\frac 1 2 m_2v_2^2$$
 
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.
 
Thank you guys, and I'll fix that in the future. I ended up finding the correct answer and it was an algebra mistake.
 
Thread 'Correct statement about size of wire to produce larger extension'
The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
20
Views
2K
  • · Replies 13 ·
Replies
13
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 15 ·
Replies
15
Views
3K
Replies
3
Views
989
  • · Replies 16 ·
Replies
16
Views
4K