Two masses connected by a spring

  • Thread starter Thread starter grusini
  • Start date Start date
  • Tags Tags
    Spring Two masses
Click For Summary
SUMMARY

The discussion focuses on a physics problem involving two masses, ##m_1=2kg## and ##m_2=4kg##, connected by a spring with a constant of ##k_1=3N/m## and a second spring with a constant of ##k_2=12N/m##. The velocities of the masses when the first spring is at rest are calculated as ##v_1=-0.016m/s## and ##v_2=0.008m/s##. The maximum compression of the second spring is determined to be ##\Delta x=0.0046m##. Participants confirm discrepancies between their calculations and textbook values, suggesting the textbook may contain errors.

PREREQUISITES
  • Understanding of conservation of energy principles
  • Knowledge of conservation of momentum
  • Familiarity with potential energy calculations for springs
  • Basic algebra for solving equations
NEXT STEPS
  • Review the principles of conservation of energy in mechanical systems
  • Study the conservation of momentum in elastic collisions
  • Learn about potential energy formulas for springs, specifically Hooke's Law
  • Explore error analysis techniques in physics problem-solving
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and spring systems, as well as educators seeking to clarify concepts related to energy conservation and momentum.

grusini
Messages
7
Reaction score
0

Homework Statement


Two masses ##m_1=2kg##, ##m_2=2m_1## are placed on the opposite edges of a spring of constant ##k_1=3N/m##, compressed of a length ##x_1=1.73cm##. The system is located on a smooth plane. At the right end of the plane there is a second spring of constant ##k_2=12N/m##. Once the first spring is at rest the masses are free to move (they are not fixed to the spring). So the situation is as follows:

_________________M1 spring1 M2___________spring2 ||

a) Find the velocity ##v_1,v_2## of masses ##m_1,m_2## respectively, when the first spring is at rest.

b) Find the maximum compression ##\Delta x## of the second spring.

Homework Equations


Conservation of energy, conservation of momentum, potential energy of a spring


The Attempt at a Solution



a) The velocities must satisfy the system:
$$\begin{cases}\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2=\frac{1}{2}k_1x_1^2 \\
m_1v_1+m_2v_2=0\end{cases}$$
##\Rightarrow v_2=\sqrt{\frac{m_1k_1x_1^2}{m_2^2+m_1m_2}}=##
##\sqrt{\frac{2\cdot 3\cdot (0.0173)^2}{4^2+4\cdot 2}}=0.008 m/s##.
Then ##v_1=-\frac{m_2}{m_1}v_2=-2v_2=-0.016m/s##.

b) The value ##\Delta x## must satisfy:
##\frac{1}{2}m_2v_2^2=\frac{1}{2}k_2(\Delta x)^2 \Rightarrow \Delta x=\sqrt{\frac{m_2v_2^2}{k_2}}=\sqrt{\frac{4\cdot (0.008)^2}{12}}=0.0046m##

Values are wrong on my textbook..
 
Physics news on Phys.org
What do you mean the values are wrong in your textbook? Do you mean your textbook gives you a solution that is different from your own? What is the given solution?

Your work seems to make sense to me. Perhaps you shouldn't be rounding the value you get for v_2 when you work through the whole problem, ie pretend 0.00865 m/s is entirely significant until the end (also, you should probably round that to 0.009 if you're going to round)
 
I talked to my professor and he agrees that the values found in my textbooks are wrong (what I meant before was that my result differs from that of the textbook). Anyway, thank you Alucinator for the tip about the rounding!
 

Similar threads

Replies
5
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K