Answer 1

Given,
Speed of the block A = 10 m/s 
The block B is kept at rest.
Coefficient of friction between floor and block B, μ = 0.10

Lets v₁ and v₂ be the velocities of A and B after collision respectively.

(a) If the collision is perfectly elastic, linear momentum is conserved.

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

\[m u_1  + m u_2    =   m v_1  + m v_2 \] 

\[ \Rightarrow 10 + 0 =  v_1  +  v_2 \] 

\[           v_1  +  v_2  = 10         .  .  . \left( 1 \right)\] 

\[\text{ We  know,} \] 

\[\text{ Velocity  of  separation }  \left( \text{ after  collision }\right)   = \text{ Velocity  of  approach }  \left( \text{ before  collision } \right )\]

\[ v_1  -  v_2    =    - ( u_1  -  v_2 )\] 

\[ \Rightarrow  v_1  -  v_2  =  - 10         .  .  . \left( 2 \right)\] 

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

\[2 v_2    =   20\] 

\[ \Rightarrow    v_2  = 10  \text{ m/s }\]
The deceleration of block B is calculated as follows:

Applying the work energy principle, we get:

\[\left( \frac{1}{2} \right) \times m \times (0 )^2  - \left( \frac{1}{2} \right) \times m \times  v^2    =    - m \times a \times  s_1 \] 

\[ \Rightarrow    - \left( \frac{1}{2} \right) \times (10 )^2    =    - \mu g \times  s_1 \] 

\[ \Rightarrow  s_1    =   \frac{100}{2 \times 1 \times 10}   =   50 \text{ m }\]

(b) If the collision is perfectly inelastic, we can write:

\[m \times  u_1  + m \times  u_2    =   (m + m) \times v\] 

\[ \Rightarrow   m \times 10 + m \times 0   =   2m \times v\] 

\[ \Rightarrow   v   =   \left( \frac{10}{2} \right)   =   5 \text{ m/s }\]
The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:

\[\left( \frac{1}{2} \right) \times 2m \times (0 )^2    -   \left( \frac{1}{2} \right) \times 2m \times (v )^2    =   2m \times \mu g \times  s_2 \] 

\[ \Rightarrow \frac{(5 )^2}{0 . 1 \times 10 \times 2}   =    s_2 \] 

\[ \Rightarrow    s_2    =   12 . 5\]  m