Question

The friction coefficient between the horizontal surface and each of the block shown in figure is 0.20. The collision between the blocks is perfectly elastic. Find the separation between the two blocks when they come to rest. Take g = 10 m/s².

Answer 1

Given:
Initial velocity of 2 kg block, v₁ = 1.0 m/s
Initial velocity of the 4 kg block, v₂ = 0 

Let the velocity of 2 kg block, just before the collision be u₁.

Using the work-energy theorem on the block of 2 kg mass:

The separation between two blocks, s = 16 cm = 0.16 m

\[\therefore \left( \frac{1}{2} \right)m \times  u_1^2  - \left( \frac{1}{2} \right)m \times (1 )^2  =    - \mu \times mg \times s  \] 

\[ \Rightarrow    u_1    =   \sqrt{(1 )^2 - 2 \times 0 . 20 \times 10 \times 0 . 16}\] 

\[ \Rightarrow    u_1    =   0 . 6  \text{ m/s}\]

As the collision is perfectly elastic, linear momentum is conserved.

Let v₁, v₂ be the velocities of 2 kg and 4 kg blocks, just after collision.

Using the law of conservation of linear momentum, we can write:

\[m_1  u_1  +  m_2  u_2  =  m_1  v_1  +  m_2  v_2 \] 

\[ \Rightarrow   2 \times 0 . 6 + 4 \times 0 = 2 v_1  + 4 v_2 \] 

\[ \Rightarrow   2 v_1  + 4 v_2  = 1 . 2       .  .  . (1)\] 
For elastic collision,
Velocity of separation (after collision) = Velocity of approach (before collision)

\[i . e .    v_1  -  v_2    =    + ( u_1  -  u_2 )\] 

                    \[  =  + (0 . 6 - 0)\] 

\[ \Rightarrow  v_1  -  v_2    =    - 0 . 6       .  .  . (2)\] 

\[\text{ Substracting  equation  (2)  from  (1),   we  get: }\] 

\[3 v_2  =   1 . 2\] 

\[ \Rightarrow    v_2    =   0 . 4  \text{ m/s}\] 

\[ \therefore      v_1  =    - 0 . 6 + 0 . 4   =    - 0 . 2  \text{ m/s} \]

Let the 2 kg block covers a distance of S₁.

∴ Applying work-energy theorem for this block, when it comes to rest:

\[\left( \frac{1}{2} \right) \times 2 \times (0 )^2  +   \left( \frac{1}{2} \right) \times 2 \times (0 . 2 )^2  =    - 2 \times 0 . 2 \times 10 \times  S_1 \] 

\[ \Rightarrow    S_1  =   1  cm .\]

Let the 4 kg block covers a distance of S₂.

Applying work energy principle for this block:

\[\left( \frac{1}{2} \right) \times 4 \times (0 )^2  -   \left( \frac{1}{2} \right) \times 4 \times (0 . 4 )^2  =    - 4 \times 0 . 2 \times 10 \times  S_2 \] 

\[ \Rightarrow   2 \times 0 . 4 \times 0 . 4   =   4 \times 0 . 2 \times 10 \times  S_2 \] 

\[ \Rightarrow    S_2    =   4  \text{ cm}\]
Therefore, the distance between the 2 kg and 4 kg block is given as,
S₁ + S₂ = 1 + 4 = 5 cm