Two blocks collide with spring in-between.

Click For Summary
SUMMARY

The problem involves two blocks colliding with a spring in between on a frictionless surface. Block 1 has a mass of 2 kg and moves at 10 m/s, while Block 2 has a mass of 5 kg and moves at 3 m/s. The spring constant is 1120 N/m. The maximum compression occurs when both blocks have the same velocity, and the correct approach to solve this is through conservation of energy or momentum, rather than using differential equations. The initial attempt using F=ma was flawed due to the moving nature of the spring's other end, leading to incorrect assumptions about the forces involved.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with conservation of momentum and energy principles
  • Knowledge of spring mechanics and Hooke's Law
  • Ability to solve systems of equations in physics contexts
NEXT STEPS
  • Learn about conservation of momentum in elastic collisions
  • Study the principles of conservation of energy in mechanical systems
  • Explore the dynamics of springs in moving systems
  • Practice solving problems involving multiple bodies and springs
USEFUL FOR

Students studying physics, particularly those focusing on mechanics, as well as educators looking for examples of collision problems involving springs.

Jose Bouza
Messages
9
Reaction score
0

Homework Statement


Block 1 (mass 2 kg) is moving rightward at 10 m/s and block 2 (mass 5kg kg) is moving rightward at 3 m/s. The surface is frictionless, and a spring with spring constant of 1120 N/m is fixed on the left side on block 2. When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

Homework Equations

The Attempt at a Solution


I realize after the fact that this could probably be solved very easily with conservation of energy and/or momentum but I tried a different route and I'm confused as to why it gave me the wrong answer. What I did was use F=ma to set up a differential equation for the velocity of Block 1 in terms of its compression of the block x.
Since $$F_1 = -kx$$ the velocity at any position x is given by the solution to the equation
\frac{dv}{dx}m = -kx
which is
$$v_1(x) = v_i1 -\frac{kx^2}{2m_1}$$
By Newtons third law the velocity of the second block at any position x should thus be
$$v_2(x) = v_i2 +\frac{kx^2}{2m_2}$$
If we set these equal to each other, plug in the relevant variables and solve for x we get a solution x = 0.1336m. When checked against the solution manual this is wrong though, where did I go wrong? Thanks.
 
Physics news on Phys.org
Hello Jose,

I think it goes wrong at the very first ##
F_1 = -kx##. This is true for a spring with the other end fixed, but here the other end moves, so you have something with ##
F_1 = -k(x_1 - x_2 - C) ## with C something like ##x_1-x_2## when both blocks touch the end of the spring (without compressing it).
 
BvU said:
Hello Jose,

I think it goes wrong at the very first ##
F_1 = -kx##. This is true for a spring with the other end fixed, but here the other end moves, so you have something with ##
F_1 = -k(x_1 - x_2 - C) ## with C something like ##x_1-x_2## when both blocks touch the end of the spring (without compressing it).
Yup you're right. When I add this into my solution I end up with a system of equations that I can't seem to solve for x2-x1 so I guess the energy method is the only one that works here. Thanks anyways.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
8
Views
6K
  • · Replies 37 ·
2
Replies
37
Views
2K
Replies
16
Views
3K
Replies
29
Views
3K
Replies
17
Views
2K
Replies
2
Views
2K
Replies
3
Views
2K