Question

Two steel balls fall freely on an elastic slab. The first ball is dropped from a height \[{h_1} = 44\,cm\] and the second from a height \[{h_2} = 11\,cm\] \[\tau s\] after the first ball. After the passage of time \[\tau \], the velocities of the balls coincide in magnitude and direction. Determine the time \[\tau \] and the time interval during which the velocities of the two balls will be equal assuming that the balls do not collide.

Answer 1

Hint: Learn about the motion of a body in free fall or under the force of gravitational acceleration. Learn about elastic collision. The velocity of a body under freefall is the product of the gravitational acceleration times the time of fall.

Formula used:
The velocity of a body falling under the influence of gravitational acceleration is given by,
\[v = gt\]
where \[g\] is the gravitational acceleration and \[t\] is the time of fall.

Complete step by step answer:
Here we have two balls that are released from a certain height after a time interval \[\tau \] from a height \[{h_1} = 44\,cm\] and \[{h_2} = 11\,cm\] respectively. Now, we know that the velocity of a body falling under the influence of gravitational acceleration is given by,
\[v = gt\] where \[g\] is the gravitational acceleration and \[t\] is the time of fall.

So, we can see that the velocity at an instant is dependent on the time but is not related to the height of the object in any way. Though the final velocity of the object does depend on the height of the object but the velocity of the object at any instant will not depend on the object.

Now, when the first ball is dropped the second ball is dropped after a time interval \[\tau \]. So, the first ball will always lead the second ball by a velocity of \[g\tau \] until it hits the ground and bounces back. Now, after impact until the second ball reaches the ground the direction of velocity will be opposite to the other .Hence, it will not be equal for any part of it. Now, after the impact the same case happens like the free fall the velocity decreases by the same factor \[gt\].

So, also in the rebound motion the velocities of the balls will not be equal in any moment of time. And this process will go on a loop until both of them stop. So, the problem given here is unsolvable as the velocities of the balls will never be equal in magnitude and direction at any moment. So, they never coincide in magnitude and direction.

Note: The velocity of the balls when it bounces back is given by, \[v = \sqrt {2gh} \] where is the height of the balls at where they were released. After the rebound the velocity decreases as \[v = \sqrt {2gh} - gt\] and stops at the maximum height.