Question

Identical balls are dropped from the same initial height and bounce back to half the initial height. In Case 1 the ball bounces off a cement floor and in case 2 the ball bounces off a piece of stretchy rubber. 1)In which case is the average force acting on the ball during the collision the biggest?

Answer 1

Answer:

Case 1

Explanation:

Given :Identical balls are dropped from the same initial height and bounce back to half the initial height

In case 1 where the the ball bounces off a cement floor, because in this case the time of contact will be less( as compared with time of contact with the stretchy rubber) And hence, a greater force will be felt by the ball ( actually impulse will be more).

impulse I= ΔF/Δt

ΔF= change in force and Δt =change in time

Answer 2

The average force acting on case 1 (cement floor) will be the greatest because of short time of contact.

The average force experienced in each case is given by Newton's second law of motion;

$F = ma = \frac{m(v_2 - v_1)}{t}$

where;

• m is the mass of the object

• v₁ and v₂ are the initial and final velocity respectively.

• t is the time of impact

The magnitude of the force is inversely proportional to the time of impact of the object.

In case 1:

• the cement floor is a hard surface and the time spent by the ball before bouncing off will be short.

In case 2:

• the stretchy rubber is elastic and the time spent by the ball before bouncing off will be greater compared to the cement.

Thus, the average force acting on case 1 (cement floor) will be the greatest because of short time of contact.

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