Question

The left block in figure (12−E13) moves at a speed v towards the right block placed in equilibrium. All collisions to take place are elastic and the surfaces are frictionless. Show that the motions of the two blocks are periodic. Find the time period of these periodic motions. Neglect the widths of the blocks.
Figure

Answer 1

The time period of these periodic motions is $2\left(\frac{L}{V}\right)+\pi \sqrt{\frac{m}{k}}$

Explanation:

In a elastic collision, when block m moves with velocity v and collides with block n, then all energy is transferred to block n. The block m moves 'x' distance with block n and block n comes to its original position.

Time period of block n is:

$T=\frac{2 \pi \sqrt{\frac{m}{k}}}{2}=\pi \sqrt{\frac{m}{k}}$

When block n collides with block m, block m comes to rest and block n moves some distance to come to its original position.

So, the periodic motion of the blocks is:

$m \rightarrow n \ \& \ n \rightarrow m$

Now, the time period is:

$\Rightarrow \frac{L}{V}+\frac{L}{V}=2\left(\frac{L}{V}\right)$

Thus, the time period of the periodic motion is:

$2\left(\frac{L}{V}\right)+\pi \sqrt{\frac{m}{k}}$

Answer 2

To show that the motions of the blocks are periodic.

Explanation:

• As all the collisions are elastic, when the block A collides with block B, block A transfers all of the energy to block B.

• When the block A moves a distance x away from the spring, block B will return to the original point and completes half of the oscillation.  Thus the blocks move periodically.  
• To find the time period of these periodic motions:
• The block B time period is $=\pi \sqrt{\frac{m}{k}}$. When the block B collides with block A and comes to rest, block A moves a distance L to return to its original position.
• Thus, the time taken is $=2\left(\frac{L}{v}\right)$.  Hence, the time period of the periodic motion of the blocks A and B is $2\left(\frac{L}{v}\right)+\pi \sqrt{\frac{m}{k}}$.