A ball moving on a smooth horizontal table hits a rough vertical wall, the coefficient of restitution between ball and wall being $$\frac{1}{3}$$. The ball rebounds at the same angle. The fraction of its kinetic energy lost is A. $$\frac{1}{3}$$ B. $$\frac{2}{3}$$ C. $$\frac{1}{9}$$ D. $$\frac{8}{9}$$

$$rac{1}{3}$$
$$rac{2}{3}$$
$$rac{1}{9}$$
$$rac{8}{9}$$

The correct answer is $\boxed{\frac{1}{9}}$.

The coefficient of restitution is a measure of how much energy is lost when two objects collide. A coefficient of restitution of $\frac{1}{3}$ means that the ball rebounds with $\frac{1}{3}$ of the energy it had before the collision.

The kinetic energy of an object is given by the equation $KE = \frac{1}{2}mv^2$, where $m$ is the mass of the object and $v$ is its velocity.

When the ball hits the wall, its velocity changes from $v_i$ to $v_f$. The change in kinetic energy is given by the equation $\Delta KE = \frac{1}{2}m(v_f^2 – v_i^2)$.

The coefficient of restitution is defined as the ratio of the final velocity to the initial velocity, so $e = \frac{v_f}{v_i}$.

Substituting this into the equation for the change in kinetic energy, we get $\Delta KE = \frac{1}{2}m(v_f^2 – v_i^2) = \frac{1}{2}m(v_f^2 – e^2v_i^2)$.

The fraction of the kinetic energy lost is given by $\frac{\Delta KE}{KE} = \frac{1}{2}m(v_f^2 – e^2v_i^2) \div \frac{1}{2}m(v_i^2)$, which simplifies to $1 – e^2$.

In this case, the coefficient of restitution is $\frac{1}{3}$, so the fraction of the kinetic energy lost is $1 – \left(\frac{1}{3}\right)^2 = \boxed{\frac{1}{9}}$.

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