Why do we need to include impulse by string?

Question

There is this paragraph in the solution of the question :

Since ball A is suspended by an inextensible string, therefore, just after collision, it can move along horizontal direction only. Hence, a vertically upward impulse is exerted by thread on the ball A. This means that during collision two impulses act on ball A simultaneously. One is impulse interaction J between the balls and the other is impulsive reaction J’ of the thread.

But, why do we need to include impulse by thread? Can't we just apply concept of conservation of mechanical energy? I mean, it got horizontal velocity and this horizontal velocity will act as tangential velocity and will help ball to complete one vertical circle. But, the book doesn't seem to understand this, or am I missing something?

Answer 1

If there is no impulse via the thread and the collision between A and B is elastic then the outcome of the collision cannot be as described : ball A cannot move horizontally.

Proof :

Linear momentum must be conserved in the x and y directions. If A moves horizontally after the collision then B must also have a horizontal component of velocity which it did not have before. Ball A has no vertical velocity after the collision so the vertical component of B's velocity must be the same after the collision as it was before.

B's total velocity and therefore its kinetic energy has increased, and A also has some kinetic energy after the collision. The balls have more kinetic energy after the collision than they had before. But this contradicts the assumption that the collision was elastic - ie that kinetic energy is conserved by the collision.

Conclusion :

Some other impulsive force must have provided the extra kinetic energy. This impulse must have come from the string.

Answer 2

Since the ball(falling) collides with the hanging ball it will give the hanging ball an impulse $I_1$which will have two components one along the vertical direction and one along the horizontal direction and so the ball will tend to go downward but as it is connected to a string which is ideal so the string will have to give an impulse of $I_2$ which is equal to the component of the impulse given by ball(falling) in vertical direction