One sphere collides with another sphere of same mass at rest inelastically. If the value of coefficient of restitution is 1/2 , the ratio of their speeds after collision shall be

The correct option is 1:3.

Step-by-step Explanation:

Let the mass of both spheres be $m$ (since they have the same mass).
Let the initial velocity of the first sphere be $u$, and the second sphere is initially at rest, so its initial velocity is $0$.
Let $v_1$ and $v_2$ be the velocities of the first and second spheres after the collision, respectively.

1. Using the Law of Conservation of Linear Momentum:
The total momentum before the collision must equal the total momentum after the collision:

$mu+m\left(0\right)=m{v}_1+m{v}_2$

Dividing the entire equation by the common mass $m$, we get:

$u=v_1+v_2$     --- (Equation 1)

2. Using the Definition of Coefficient of Restitution ($e$):
The coefficient of restitution is defined as the ratio of the relative velocity of separation after collision to the relative velocity of approach before collision:

$e=\frac{v_2-v_1}{u-0}$

Given that $e=\frac{1}{2}$, we substitute this value into the equation:

$\frac{1}{2}=\frac{v_2-v_1}{u}$

Rearranging the terms gives:

$v_2-v_1=\frac{u}{2}$     --- (Equation 2)

3. Solving for the Final Velocities:
Adding Equation 1 and Equation 2:

$(v_1+v_2)+(v_2-v_1)=u+\frac{u}{2}$

$2v_2=\frac{3u}{2}$

$v_2=\frac{3u}{4}$

Now, substitute the value of $v_2$ back into Equation 1 to find $v_1$:

$v_1=u-v_2$

$v_1=u-\frac{3u}{4}$

$v_1=\frac{u}{4}$

4. Finding the Ratio of Their Speeds After Collision:
The ratio of their speeds ($v_1:v_2$) is:

$\frac{v_1}{v_2}=\frac{\left(\frac{u}{4}\right)}{\left(\frac{3u}{4}\right)}=\frac{1}{3}$

Therefore, the ratio of their speeds after collision is 1:3.