Question

In the figure shown, a block $A$ moving with velocity $20\text{m/s}$ on a horizontal surface collides with another identical block $B$, initially at rest. The coefficient of restitution is $0.5$. Neglect friction everywhere. The distance between the blocks, $5\text{sec}$ after the collision takes place is

• A. $50\text{m}$
• B. $125\text{m}$
• C. $100\text{m}$
• D. $150\text{m}$

Answer 1

The correct option is A $50\text{m}$

Let mass of each block be $m$.
$u_1=20\text{m/s}$ (initial velocity of block $A$ before collision)
$u_2=0\text{m/s}$ (initial velocity of block $B$ before collision)
Let $v_1=$ final velocity of block $A$ after collision
$v_2=$ final velocity of block $B$ after collision

There is no external force on the system, hence linear momentum will be conserved.
Applying conservation of momentum,
$m\times 20+m\left(0\right)=m{v}_1+m{v}_2$
$\Rightarrow v_1+v_2=20$ ... ($1$)

As we know,
$e=\frac{v_2-v_1}{u_1-u_2}$
$\Rightarrow 0.5=\frac{v_2-v_1}{20-0}$
$\Rightarrow v_2-v_1=10$ ... ($2$)

From ($1$) and ($2$) equation,
$2v_2=30$
$v_2=15\text{m/s}$ and $v_1=5\text{m/s}$

Distance between two blocks$\left(s\right)$ after the collision at $t=5\text{s}$
$s=(v_2-v_1)\times t$
$\Rightarrow s=(15-5)\times 5=50\text{m}$
Hence, the correct option is (a)