Welcome to Physics Problems Q&A, where you can ask questions and receive answers from other members of the community.

Completly Inelastic Collision in a Spring Block System

2 votes
729 views

2 blocks A and B of masses $m$ and $2m$ respectively are connected by a spring of spring constant K the masses are moving to the right with an uniform velocity $v_0$ each, the heavier mass leading the lighter one. The spring has its natural length during this motion block B collides head on with a 3rd Block C of mass $2m$ which was initially at rest.

  1. What is the velocity of B just after the collision ?
  2. What is the velocity of the center of mass?
  3. What is the maximum compression of the spring after the collision?

What so far I concluded is that I thought of using the reduced mass concept and reduced mass $\frac {2}{3}$m. $v_0$ =-$\frac {8}{3}mv$ where $v$ is the velocity of the center of mass.

But I am not so confident that I am right because I am unable to identify whether the external force is 0 or not. Moreover is it okay to use reduced mass concept? and how to think about the length of spring and its calculation?

asked Nov 24, 2016 in Physics Problems by physicsapproval (2,290 points)
edited Nov 25, 2016 by Einstein

1 Answer

1 vote
 
Best answer

At the time of the collision between B and C (which we must assume is instantaneous), the only force on B is that due to C. There is no force on B from the spring, because the spring is at its natural length at this instant. So during the collision between B and C you can ignore the spring and A. Using reduced mass (which takes account of A) will give you the wrong answer.

Is C initially stationary? The question - as you have posted it - does not say. Your title says the collision is completely inelastic, but that is not mentioned in the text either. Have you missed out some information?

Finding the velocity of the CM of A and B after the collision is useful because in the CM frame the 2 objects A and B come to rest when the spring reaches maximum compression : the PE stored at maximum compression equals the KE of A and B.


Assuming that B and C merge on collision, their final mass is $4m$ so to conserve momentum their velocity must be $\frac12 v_0$.

The velocity of the CM of the system ABC is $\frac{mv_0+4m\frac12v_0}{m+4m}=\frac35v_0$.

In the CM frame the velocities of A and BC are $v_0-\frac35v_0=\frac25v_0$ and $\frac12v_0-\frac35v_0=-\frac{1}{10}v_0$. The total KE is $\frac12 m ( \frac25 v_0)^2+\frac12 (4m)(-\frac{1}{10} v_0)^2=\frac{1}{10}mv_0^2$. This equals the energy stored in the spring at maximum compression, so
$\frac12Kx^2=\frac{1}{10}mv_0^2$
$x^2=\frac{mv_0^2}{5K}$
$x=v_0\sqrt{\frac{m}{5K}}$.

answered Nov 24, 2016 by sammy gerbil (26,660 points)
edited Nov 27, 2016 by sammy gerbil
Why did we take kinetic energy of the relatove velocities and not of their original velocities to find maximum compression?
I used the centre of momentum (CM) frame of reference to make this calculation easier. In this frame when the spring has its maximum compression then the velocities of all 3 blocks A, B and C are all zero. There is no longer any KE, so the initial KE when there was zero compression of the spring has been fully transformed into elastic PE stored in the spring.

If we used the velocities relative to the ground then at maximum spring compression the blocks would still have some velocity relative to the ground, they would still  have some KE. We could not say that the initial KE of the blocks equals the elastic PE stored in the spring at maximum compression.  

It is possible to do the same calculation using velocities relative to the ground, and  to get the correct answer. However, using the CM frame of reference makes the calculation easier.
...