Find V(f) for two particles resting (and compressing) against a spring.

In summary, two particles of mass m1= 2.2 kg and m2 = 4.5 kg compress a spring with a spring constant k = 395 N/m by 0.15 m on a horizontal track. Using conservation of momentum and energy, the final velocities of the particles are calculated to be 1.17 m/s and 0.611 m/s respectively. However, there may be an error in the values used for calculation.
  • #1
Covenant32
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0

Homework Statement


Two particles of mass m1= 2.2 kg and mass m2 = 4.5 kg that are free to move on a horizontal track are initially held at rest so that they compress a spring as shown in the figure below. The spring has a spring constant k = 395 N/m and is compressed 0.15 m. Find the final velocities of the two particles.


Homework Equations



M1V1=M2V2

1/2M1V1^2 + 1/2M2V2^2 = 1/2 kx^2


The Attempt at a Solution



I am copy/paste a solution for the problem that is exactly what I did but with different numbers.

Conservation of momentum:
m1 v1 + m2 v2 = 0 ...(1)

Conservation of energy:
m1 v1^2 / 2 + m2 v2^2 / 2 = kx^2 / 2
m1 v1^2 + m2 v2^2 = kx^2 ...(2)

From (1):
v2 = - m1 v1 / m2

Substituting for v2 in (2):
m1 v1^2 + m1^2 v1^2 / m2 = kx^2

m1 m2 v1^2 + m1^2 v1^2 = k m2 x^2
m1(m1 + m2)v1^2 = k m2 x^2
v1^2 = k m2 x^2 / [ m1(m1 + m2) ]

v1 = sqrt{ 395 * 4.4 * 0.11^2 / [ 2.3(2.3 + 4.4) ] }
v1 = 1.17 m/s to 3 sig. fig.

Similarly:
v2^2 = k m1 x^2 / [ m2(m1 + m2) ]

v2 = sqrt{ 395 * 2.3 * 0.11^2 / [ 4.4(2.3 + 4.4) ] }
v2 = 0.611 m/s to 3 sig. fig.


I did this and my answer is incorrect.


I ALSO tried:

0.5*k*x^2 = 0.5*m1*v1^2
0.5*k*x^2 = 0.5*m2*v2^2

and THAT was incorrect. I'm in the dark here.


By the way, this is my first post, though I've been browsing PF all semester! Thanks for all the previous help =)
 
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  • #2
Oh, I forgot to speak of the figure:

0~~~O Two figures are compressing a spring .15 m.
 
  • #3
Your algebra looks good to me. Check the part where you plug in values, though.
 

Related to Find V(f) for two particles resting (and compressing) against a spring.

1. What is V(f) for two particles resting against a spring?

V(f) stands for the potential energy of the system of two particles resting against a spring. It represents the energy stored in the spring due to its deformation from its equilibrium position.

2. How is V(f) calculated for a system with two particles and a spring?

V(f) can be calculated using the equation V(f) = 1/2kx^2, where k is the spring constant and x is the displacement of the spring from its equilibrium position due to the weight of the particles.

3. Can V(f) be negative for two particles resting against a spring?

Yes, V(f) can be negative if the displacement of the spring is in the opposite direction of the force applied by the particles. This means that the system has lost potential energy and it has been converted into kinetic energy.

4. How does the mass of the particles affect V(f) in this system?

The mass of the particles does not directly affect V(f) in this system. However, it does affect the displacement of the spring and therefore the potential energy stored in the spring.

5. Is V(f) constant in this system or does it change over time?

V(f) is not constant in this system. As the particles compress the spring, the displacement and therefore the potential energy of the spring will change. However, if no external forces are applied, the potential energy will remain constant at its maximum value when the particles are at their resting position.

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