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[Dan350]

I think I almost got this problem would you review or correct me please?:

A spring with a spring constant k=34/Nm is attached to a block of mass M1=1.5kg. Initially that block is at rest on a friction less surface. A second block, M2, of mass of 4.5kg is moving with speed v= 5.5m/s toward the first block. When the second block hits the first block, the two blocks slide together to the left, compressing the spring, and then rebound for a half cycle of harmonic motion.

You can see an image below*

a) With what speed does M2 leave M1 after the half cycle of oscillation?

b) What time is takes for the half period of simple harmonic motion during which the block are in contact?

c)What is the maximum compression of the spring
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Relevant Equations

Pi=Pf
Δp=FΔt

K+PE=ΔU
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My attempt
a)
Half cycle of oscillation is at the moment the spring is at its eqilibriuim after the collision( that's what i think)

for a I used conservation of momentum ∴ P1=Pf

so my attempt: m2Vi=(m1+m2)vf
where vi= initial velocity and vf= final velocity

Solving for vf i get 4.125

b)
Since the momentum is going both negative and positive durin the half cycle
Δp=fΔt
-p-p=fΔt
-2(m1+m2)vf=-kxt

solving for t i got 84 seconds

c) the x from above I got it from here;

1/2(m1+m2)vf^2=1/2(kx^2)

solving for x i got 1.73m
Any corrections?
Thank you in advance

Cheers

-Daniel

 

[haruspex]

A spring with a spring constant k=34/Nm

You mean 34N/m.

so my attempt: m2Vi=(m1+m2)vf
where vi= initial velocity and vf= final velocity

Solving for vf i get 4.125

That is the right answer for a), but you have not competed the explanation. It looks like you have only calculated the velocity immediately after collision. You need to explain why this is also M2's velocity when they later separate.

b)
Since the momentum is going both negative and positive durin the half cycle
Δp=fΔt
-p-p=fΔt
-2(m1+m2)vf=-kxt

solving for t i got 84 seconds

That would work if the force were a constant, but it increases as the spring compresses: Δp=∫f.dt

 

[hjelmgart]

The question seems a bit strange to me. Is this a translation of some kind?

a)
A full cycle of oscillation means, it ends up back, where it started. So half would of course not make it end in its equilibrium position again.

But the way the question is formulated, they just want to know the speed of M2, when it leaves M1. Is this correct? So techniqually it seems like you did the right thing with your math, anyway.


b) I am not so sure about your method in this one, but think about how the period of such a spring is defined:
T = 2 pi/ω and ω = √(k/m)
The time you need is at half the full cycle, meaning t = T/2

c)
You can use the time and the starting velocity to find the acceleration from when the velocity goes from vf to zero.

Then you can find the compression using Hooke's law:

F=-k*x = m*a and solve for x!


Although I must say, I am not that sure, if what you did is wrong, but you can always try this method, and see what it gives. :-)
At least when the questions are in such order, I don't think you should find the compression, before you are asked to!

 

[haruspex]

The question seems a bit strange to me. Is this a translation of some kind?

a)
A full cycle of oscillation means, it ends up back, where it started. So half would of course not make it end in its equilibrium position again.

The impact and subsequent separation will both be at the spring's equilibrium position, but are only separated by half a cycle.

 

[hjelmgart]

The impact and subsequent separation will both be at the spring's equilibrium position, but are only separated by half a cycle.

Ah, you are of course correct. A full cycle would mean, that the block is moving in -x direction again, as it would have to be fully stretched as well. The block leaves before that happens, namely at equilibrium, where the force in the spring starts to create a deacceleration.

 

[Dan350]

You mean 34N/m.

That is the right answer for a), but you have not competed the explanation. It looks like you have only calculated the velocity immediately after collision. You need to explain why this is also M2's velocity when they later separate.

That would work if the force were a constant, but it increases as the spring compresses: Δp=∫f.dt

So what would be my limits of integration then? Or how you approach this problem?

Thanks

 

[Dan350]

The question seems a bit strange to me. Is this a translation of some kind?

a)
A full cycle of oscillation means, it ends up back, where it started. So half would of course not make it end in its equilibrium position again.

But the way the question is formulated, they just want to know the speed of M2, when it leaves M1. Is this correct? So techniqually it seems like you did the right thing with your math, anyway.


b) I am not so sure about your method in this one, but think about how the period of such a spring is defined:
T = 2 pi/ω and ω = √(k/m)
The time you need is at half the full cycle, meaning t = T/2

c)
You can use the time and the starting velocity to find the acceleration from when the velocity goes from vf to zero.

Then you can find the compression using Hooke's law:

F=-k*x = m*a and solve for x!


Although I must say, I am not that sure, if what you did is wrong, but you can always try this method, and see what it gives. :-)
At least when the questions are in such order, I don't think you should find the compression, before you are asked to!

Thank you, so for b)


T/2= 2pi/√(k/m) right?

And m for c). Will be m1+m2 right?
Thanks

 

[hjelmgart]

Thank you, so for b)


T/2= 2pi/√(k/m) right?

And m for c). Will be m1+m2 right?
Thanks

You divide by 2 on both sides in b), so

t = T/2 = pi/√(k/m)


and yes m = m1+m2 in c)

 

[Dan350]

You divide by 2 on both sides in b), so

t = T/2 = pi/√(k/m)


and yes m = m1+m2 in c)

Yess,, you were right on that one,thanks,, and finally

x=ma/k right?

just solving for it