Two Body Spring Problem

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  • #1
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Homework Statement


Two blocks sitting on a frictionless horizontal surface are connected by a light spring. The natural length of the spring is ##d##, and the spring constant is ##k##. The two objects are initially both at rest, with the spring neither stretched nor compressed. At time ##t = 0##, ##m_1## is struck a blow that gives it an initial velocity ##v_0## toward ##m_2##. Taking the initial position of the centre of mass to be zero, use the equations of motion for ##\ddot R## and ##\ddot r## to show that the center of mass and relative position are given by

##R=\frac{m_1v_0}{M}t##, ##r=C+Asin(\omega t+\delta)##

where ##\omega = \sqrt{k/\mu}##. Use the initial condition for ##r## to show ##\delta = 0## and ##C=d##.

Homework Equations


##m_1\ddot r_1=F##
##m_2\ddot r_2=-F##

##M\ddot R=0##
##\mu\ddot r=F##

##r_1=R+\frac{m_2}{M}r##

##r_2=R-\frac{m_1}{M}r##

The Attempt at a Solution



So ##\mu\ddot r=F## is the same as ##\mu\ddot r=m_1a_1##. Differentiating to get:

##\mu\dot r=m_1v_1##, and again

##\mu r=m_1r_1##, where ##r_1=v_0t##

Therefore ##\mu r=m_1v_0t##

##\mu r## simplifies to ##MR##. Subbing in and I get required:

##R=\frac{m_1v_0}{M}t##

I have no idea how to get ##r=C+Asin(\omega t+\delta)## however. Any help would be greatly appreciated.
 

Answers and Replies

  • #2
DEvens
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Ok, I think you are having some conceptual difficulty here. Do you understand the following?

- ##R## is the location of the centre of mass of the system.
- ##r## is the distance from one mass to the other.

So at the start, before the impact, everything is at rest. So at the start there is no momentum for anything. So ##R## starts out as 0. And ##r## starts out as ##d##. So far so good?

After the impact there is some momentum. This means that the centre of mass of the system has to be moving. But after the impact there are no more external forces. So after the impact, the centre of mass of the system will move with constant velocity. So what does that mean for ##R##?

Now to get the equation for ##r## the easy thing is to go to the centre of mass system of coordinates. In that system you have the centre of momentum at rest, and ##R'## is just 0. In that system, just after the collision, you have the two masses moving towards each other such that they have equal magnitude but opposite sign momentums.

Can you work it out now?
 
  • #3
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After the impact there is some momentum. This means that the centre of mass of the system has to be moving. But after the impact there are no more external forces. So after the impact, the centre of mass of the system will move with constant velocity. So what does that mean for RR?
I assume that means that ##R## changes with respect to time.

Now to get the equation for rr the easy thing is to go to the centre of mass system of coordinates. In that system you have the centre of momentum at rest, and R′R' is just 0. In that system, just after the collision, you have the two masses moving towards each other such that they have equal magnitude but opposite sign momentums.

Is this the conservation of momentum?
 
  • #4
haruspex
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I assume that means that ##R## changes with respect to time.
Yes, but in what way, exactly? Can you write an equation involving ##\ddot R## from that?
Is this the conservation of momentum?
Not really. DEvens is advising you to use the frame of reference of the common mass centre. The behaviour of that common mass centre, R, discussed above, is governed by conservation of momentum. DEvens' later remark about how the two masses move in relation to each other within that reference frame follows from the definition of mass centre.
 

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