Block collision attached to spring

[naianator]

Homework Statement

A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.

Homework Equations

U_spring = 1/2*k*x^2

K = 1/2*m*v^2

F = m*a

v_f^2 = v_0^2+2*a*x

The Attempt at a Solution

I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x. I'm really not having any luck.

E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

at x/2 where m collides with 2m:

1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

k*x^2 = 3m*v^2 + 1/4*k*x^2

3/4*k*x^2 = 3*m*v^2

yielding

v = sqrt(k*x^2/(4*m))

by Newtons second law a = -k*x/(3*m)

and using kinematics (v_f^2 = v_0^2+2*a*x)

0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

x/4 = 2/3*delta(x)

delta(x) = 3/8*x

Am I way off base with this approach?

 

[Doc Al]

I used conservation of energy to find the velocity where the two blocks collide at x/2.

You can use conservation of energy to find the velocity of mass m just before the collision. But energy is not conserved during the collision itself. (But what is conserved?)

 

[naianator]

You can use conservation of energy to find the velocity of mass m just before the collision. But energy is not conserved during the collision itself. (But what is conserved?)

momentum?

 

[Ellispson]

Homework Statement

A block of mass m is attached to a spring with a force constant k, as in the above diagram. Initially, the spring is compressed a distance x from the equilibrium and the block is held at rest. Another block, of mass 2m, is placed a distance x/2 from the equilibrium as shown. After the spring is released, the blocks collide inelastically and slide together. How far (Δx) would the blocks slide beyond the collision point? Neglect friction between the blocks and the horizontal surface.

Homework Equations

U_spring = 1/2*k*x^2

K = 1/2*m*v^2

F = m*a

v_f^2 = v_0^2+2*a*x

The Attempt at a Solution

I used conservation of energy to find the velocity where the two blocks collide at x/2. I'm not sure about this though, because I don't think that conservation of momentum applies and I used 3m for the mass (assuming that the blocks had collided). Then I used N2L to find the acceleration based on the force (-kx) and mass (3m). I plugged that into the kinematics equation for final velocity and solved for delta x. I'm really not having any luck.

E_i = 1/2*k*x^2 = E_f = 1/2*m*v^2 + 1/2*k*x_f^2

at x/2 where m collides with 2m:

1/2*k*x^2 = 1/2*3m*v^2 + 1/8*k*x^2

k*x^2 = 3m*v^2 + 1/4*k*x^2

3/4*k*x^2 = 3*m*v^2

yielding

v = sqrt(k*x^2/(4*m))

by Newtons second law a = -k*x/(3*m)

and using kinematics (v_f^2 = v_0^2+2*a*x)

0 = k*x^2/(4*m) - 2*(k*x/(3*m))*delta(x)

x/4 = 2/3*delta(x)

delta(x) = 3/8*x

Am I way off base with this approach?

According to what I did understand of your method to solve the problem.You have conserved energy in an inelastic collision which I don't think is valid.
Conservation of linear momentum however will hold true.I suggest you use that to find the final velocities of the blocks.

Edit:I'm sorry,I didn't see the earlier replies to the post.

 

[Doc Al]

momentum?

Yes!

 

[naianator]

According to what I did understand of your method to solve the problem.You have conserved energy in an inelastic collision which I don't think is valid.
Conservation of linear momentum however will hold true.I suggest you use that to find the final velocities of the blocks.

so according to conservation of momentum mass m has velocity sqrt(3*k*x^2/(4*m)) before the collision and according to conservation of momentum the blocks have velocity sqrt(3*k*x^2/(4*m))/3 after the collision. Can I find the acceleration and distance traveled with the same approach I tried before?

 

[naianator]

Yes!

so if I find the velocity of the masses after the collision can I use N2L and kinematics to find distance traveled?

 

[naianator]

so if I find the velocity of the masses after the collision can I use N2L and kinematics to find distance traveled?

I'm assuming not because I got another wrong answer

 

[Doc Al]

so according to conservation of momentum mass m has velocity sqrt(3*k*x^2/(4*m)) before the collision

I believe you mean conservation of energy here.

and according to conservation of momentum the blocks have velocity sqrt(3*k*x^2/(4*m))/3 after the collision.

Right.

Can I find the acceleration and distance traveled with the same approach I tried before?

No. That method was flawed since you assumed constant acceleration.

During the collision, energy is not conserved. But before and after the collision it is.

so if I find the velocity of the masses after the collision can I use N2L and kinematics to find distance traveled?

No, but see above for a hint.

 

[naianator]

I believe you mean conservation of energy here.Right.No. That method was flawed since you assumed constant acceleration.

During the collision, energy is not conserved. But before and after the collision it is.No, but see above for a hint.

ahhh so I can use conservation of energy again

 

[naianator]

ahhh so I can use conservation of energy again

so is it 1/2*m*v^2 + 1/2*k*(x/2)^2 = 1/2*k*delta(x)^2

= x^2/12+x^2/4 = delta(x)^2

delta(x) = .57735?

 

[naianator]

I believe you mean conservation of energy here.Right.No. That method was flawed since you assumed constant acceleration.

During the collision, energy is not conserved. But before and after the collision it is.No, but see above for a hint.

so is it 1/2*m*v^2 + 1/2*k*(x/2)^2 = 1/2*k*delta(x)^2

= x^2/12+x^2/4 = delta(x)^2

delta(x) = .57735?

Nope. So what am I doing wrong now?

 

[Doc Al]

so is it 1/2*m*v^2 + 1/2*k*(x/2)^2 = 1/2*k*delta(x)^2

Make sure you have the right mass and speed for the KE term. And express the final PE term in terms of Δx properly. Δx is the distance beyond the collision point, not the final distance from equilibrium.

 

[naianator]

Make sure you have the right mass and speed for the KE term. And express the final PE term in terms of Δx properly. Δx is the distance beyond the collision point, not the final distance from equilibrium.

Got it, thanks.