Question

The blocks shown in figure have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.10 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic.
Figure

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:
$$\begin{gathered} m{u}_1+m{u}_2=m{v}_1+m{v}_2 \\ \Rightarrow 10+0=v_1+v_2 \\ {v}_1+v_2=10...\left(1\right) \\ \mathrm{We}\mathrm{know}, \\ \mathrm{Velocity}\mathrm{of}\mathrm{separation}\left(\mathrm{after}\mathrm{collision}\right)=\mathrm{Velocity}\mathrm{of}\mathrm{approach}\left(\mathrm{before}\mathrm{collision}\right) \\ {v}_1-v_2=-(u_1-v_2) \\ \Rightarrow v_1-v_2=-10...\left(2\right) \\ \mathrm{Substracting}\mathrm{equation}\left(2\right)\mathrm{from}\left(1\right),\mathrm{we}\mathrm{get}: \\ 2v_2=20 \\ \Rightarrow v_2=10\mathrm{m}/\mathrm{s} \end{gathered}$$

The deceleration of block B is calculated as follows:

Applying the work energy principle, we get:

$$\begin{gathered} \left(\frac{1}{2}\right)\times m\times (0{)}^2-\left(\frac{1}{2}\right)\times m\times v^2=-m\times a\times s_1 \\ \Rightarrow -\left(\frac{1}{2}\right)\times (10{)}^2=-\mu g\times s_1 \\ \Rightarrow s_1=\frac{100}{2\times 1\times 10}=50\mathrm{m} \end{gathered}$$

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

$$\begin{gathered} m\times u_1+m\times u_2=(m+m)\times v \\ \Rightarrow m\times 10+m\times 0=2m\times v \\ \Rightarrow v=\left(\frac{10}{2}\right)=5\mathrm{m}/\mathrm{s} \end{gathered}$$

The two blocks move together, sticking to each other.
∴ Applying the work-energy principle again, we get:

$$\begin{gathered} \left(\frac{1}{2}\right)\times 2m\times (0{)}^2-\left(\frac{1}{2}\right)\times 2m\times (v{)}^2=2\mathrm{m}\times \mathrm{\mu }g\times s_2 \\ \Rightarrow \frac{(5{)}^2}{0.1\times 10\times 2}=s_2 \\ \Rightarrow s_2=12.5\mathrm{m} \end{gathered}$$