Conservation Laws and Collisions with a Spring

[nibbles198]

Homework Statement

A first block with mass M_1 is initially moving with a speed V_o to the right and a second block with mass M_2 is at rest. The second block has a spring attached to it. The spring has negligible mass. The spring has a spring constant k. The horizontal surface is frictionless. (Note: the spring is facing the first block)

a) What quantities are conserved throughout the collision?

b) Write equations that express the conservation laws during the collision when the spring is compressed a distance X.

c) Now redo part b) and assume that the masses are both equal to M

d) Solve the equations in part c) for the speeds of the blocks V_1 and V_2 during the collision when the spring is compressed a distance X. Again assume the masses are equal.

e) Now assume that the spring reaches a maximum compression L during the collision. Plot V_1 and V_2 vs the compression of the spring, X (masses are equal).

f) After the collision what are the final speeds of the two blocks? (masses are equal)

g) Now remove the spring from the second block and consider the collision again, assuming that the two masses are equal. With the same initial speeds solve for the final speeds of the two blocks.

Homework Equations

The Attempt at a Solution

a)
Energy and Momentum

b)
Here's where I get confused (yes, this early in the problem unfortunately). I used:

.5*k*x^2 = .5*M_1*(V_o)^2 and
.5*k*x^2 = .5*M_2*(V_final)^2

Since the first mass should stop after hitting the spring and the second should gain all of the energy of the system in Kinetic energy. So these two problems are basically saying that the Potential energy of the spring = the kinetic energy of the first mass = the final kinetic energy of the second mass. But...it seems like this is wrong, since part d) asks for the speeds of the blocks V_1 and V_2 during the collision.

So I guess my real question here is, do the two equations I set up express the conservation laws during the collision when the spring is compressed a distance X?

Thank you in advance.

 

[tiny-tim]

welcome to pf!

hi nibbles198! welcome to pf!

(are you a rabbit? I'm a goldfish! )

Here's where I get confused (yes, this early in the problem unfortunately). I used:

.5*k*x^2 = .5*M_1*(V_o)^2 and
.5*k*x^2 = .5*M_2*(V_final)^2

Since the first mass should stop after hitting the spring and the second should gain all of the energy of the system in Kinetic energy. So these two problems are basically saying that the Potential energy of the spring = the kinetic energy of the first mass = the final kinetic energy of the second mass. But...it seems like this is wrong, since part d) asks for the speeds of the blocks V_1 and V_2 during the collision.

So I guess my real question here is, do the two equations I set up express the conservation laws during the collision when the spring is compressed a distance X?

(try using the X² and X₂ icons just above the Reply box )

no, since (b) asks for the conservation laws (btw, that includes momentum) "during the collision" you do need both v₁ and v₂ in the equation

 

[nibbles198]

Thanks , I am indeed a rabbit lol.

Ok so what are v₁ and v₂ here then? The velocity before the spring is compressed and the velocity once it is compressed? I guess I should have included this question in my original post because determining what v₁ and v₂ are is my real issue.

 

[tiny-tim]

no, v₁ and v₂ are the speeds of the two blocks at any time during the collision …

they'll depend on x, the amount of compression at any particular time

 

[nibbles198]

ok, so we know that

(1/2)kx² = potential energy in the spring when it is fully compressed.

If momentum is conserved, then when it is fully compressed

(1/2)M₁V_o² = (1/2)kx²

correct?

Hmm...we really haven't learned this in the class I'm in currently, but it seems like I would need to take the derivative of the velocity to be able to find the velocity for a given time. Is this along the right lines? If so, I need to do some more research on the problem because we have not yet learned how to do that...

 

[nibbles198]

thanks so much for your responses btw. I really did not expect anyone to even get back to me

 

[tiny-tim]

hi nibbles19!

ok, so we know that

(1/2)kx² = potential energy in the spring when it is fully compressed.

why do you say "fully compressed"?

that is the correct formula for PE for any amount of compression

now just put the two KEs into the equation

 

[nibbles198]

Ahhh I think I've got it.

(1/2)kx² = (1/2)M₁V₁² + (1/2)M₂V₂²

Awesome thank you! (let me know if that's not right, but I'm pretty sure it is)

 

[tiny-tim]

that's fine!

have a piece of lettuce! ​