Question

Two identical spheres $A$ and $B$ are free to move and to rotate about their centers. They are given the same impulse $J$ . The lines of action of the impulse pass through the center of $A$ and away from the center of $B$ , then
(A) $A$ and $B$ will have the same speed
(B) $B$ will have greater kinetic energy than $A$
(C) They will have the same kinetic energy, but the linear kinetic energy of $B$ will be less than that of $A$
(D) The kinetic energy of $B$ will depend on the point of impact of the impulse on $B$

Answer 1

Hint : Impulse can also be defined as the change in momentum. Here, we know the masses of both bodies are the same, hence by equating the impulse applied, we can find the relation between speed and kinetic energy.

Complete Step By Step Answer:
Here, the given spheres are identical. This means that the radius of the spheres and the mass of the spheres are the same.
Impulse can be defined as the integral of force with the time interval through which the force is applied.
Now, as we know that force can be defined as the time rate of change of momentum, we can define impulse as the change in momentum.
The given spheres are given the same impulse $J$ . Hence the change of momentum of both spheres will be equal.
Now, the spheres were stationary initially, which means the initial momentum of both spheres is equal to zero. Hence, the final momentum of both the spheres is equal which can be expressed as,
${m_A}{v_{fA}} = {m_B}{v_{fB}}$
The masses of both the spheres are equal, ${m_A} = {m_B}$
$\therefore {v_{fA}} = {v_{fB}}$
Hence, both the spheres are moving at the same speed.
Similarly, the angular impulse can be defined as the change in angular momentum.
Let us denote angular impulse by $H$ and angular momentum by $P$
$\therefore H = \Delta P$ …… $(1)$
Angular impulse can be calculated as $H = Jr$
Where, $J$ is the linear impulse and $r$ is the distance of the line of action of impulse from the center.
Angular momentum can be calculated by $P = I\omega$
Where, $I$ is the moment of inertia and $\omega$ is the angular velocity
Substituting these values in the equation $(1)$ ,
$\therefore Jr = \Delta \left( {I\omega } \right)$
As the initial angular momentum will be zero, the change in angular momentum will be equal to the final angular momentum.
$\therefore Jr = I\omega$
For the sphere $A$ , the line of action of impulse passes through the center, which implies $r = 0$
$\therefore J(0) = I\omega$
$\therefore \omega = 0$
Thus sphere $A$ is moving only with translational motion.
However, the line of action of impulse passes through the sphere $B$ at a distance of say $x$ from the center.
$\therefore Jx = I\omega$
Thus, sphere B is moving with translational as well as rotational motion. Hence, it has linear as well as angular kinetic velocity.
Thus, the kinetic energy of the sphere $B$ is greater than the kinetic energy of the sphere $A$ .
Also, the angular kinetic velocity of the sphere $B$ depends on the angular velocity which in turn depends on the distance of the line of action of impulse from the center.
Thus, the correct answers are Option $(A)$ , $(B)$ , $(D)$ .

Note :
Here, the masses of the spheres are equal and from the impulse, we can deduce that the velocity of both spheres is equal. Hence, the linear kinetic energy of both the spheres will be equal. The kinetic energy of the sphere $B$ is greater only because it also possesses angular kinetic energy.