The ratio of masses of two balls is 2 : 1 and before collision the ratio of their velocities is 1 : 2 in mutually opposite direction. After collision each ball moves in an opposite direction to its initial direction. If e = (5/6), the ratio of speed of each ball before and after collision would be

The correct option is 5/6 times.

Step-by-Step Explanation:

Let us define the parameters of the two balls involved in the collision:

Let the masses of the two balls be $m_1$ and $m_2$.

We are given the ratio of their masses as 2 : 1. Therefore, we can write:

$m_1=2m$

$m_2=m$

Let the velocities of the two balls before the collision be $u_1$ and $u_2$. The ratio of their velocities is 1 : 2, and they move in mutually opposite directions. Let us choose the direction of the first ball as positive:

$u_1=u$

$u_2=-2u$

Now, let us calculate the total initial linear momentum of the system before the collision ($P_i$):

$P_i=m_1{u}_1+m_2{u}_2$

Substituting the values:

$P_i=\left(2m\right)\left(u\right)+\left(m\right)(-2u)$

$P_i=2mu-2mu=0$

According to the law of conservation of linear momentum, the total momentum of the system must remain conserved. Therefore, the total momentum after the collision ($P_f$) must also be zero:

$P_f=m_1{v}_1+m_2{v}_2=0$

Where $v_1$ and $v_2$ are the velocities of the balls after the collision. Substituting the masses:

$2m{v}_1+m{v}_2=0$

$v_2=-2v_1$

The problem states that after the collision, each ball moves in a direction opposite to its initial direction. Since ball 1 originally moved in the positive direction, its final velocity must be negative. Let $v_1=-v$ (where $v>0$ is the final speed of ball 1). Substituting this into the momentum equation gives:

$v_2=-2(-v)=2v$

Thus, the speeds after the collision are $v$ for the first ball and $2v$ for the second ball.

The coefficient of restitution ($e$) is defined as the ratio of the relative velocity of separation to the relative velocity of approach:

$e=\frac{v_2-v_1}{u_1-u_2}$

Substituting the values of velocities:

$e=\frac{2v-(-v)}{u-(-2u)}$

$e=\frac{3v}{3u}=\frac{v}{u}$

We are given that $e=\frac{5}{6}$, so:

$\frac{v}{u}=\frac{5}{6}$

Now, let us find the ratio of speed of each ball after collision to its speed before collision:

$\frac{\text{Speed after collision}}{\text{Speed before collision}}=\frac{\left|v_1\right|}{\left|u_1\right|}=\frac{v}{u}=\frac{5}{6}$

$\frac{\text{Speed after collision}}{\text{Speed before collision}}=\frac{\left|v_2\right|}{\left|u_2\right|}=\frac{2v}{2u}=\frac{v}{u}=\frac{5}{6}$

Therefore, for both balls, the speed after collision is 5/6 times the speed before collision.