Question

2 kg steel ball is dropped straight down ont a hard horixontal florr and bounces straight up . the balls speed just before and just after impact with the floor is 10/s determine the magnitude of the impluese delivered to the floor by the steel ball?

Answer 1

Answer:

$40 \frac{kg*m}{s}$

Explanation:

Impulse-momentum theorem states that impulse is equal to the change of momentum:

$\overrightarrow{J}=\overrightarrow{p}_{f}-\overrightarrow{p}_{i}$ (1)

with pf the final momentum and pi the initial momentum. Knowing that momentum is mass (m) times velocity (1) is:

$\overrightarrow{J}=m \overrightarrow{v}_{f}- m \overrightarrow{v}_{i}$

It's important to note that we're dealing with vector quantities so direction matters. If we choose towards the floor positive direction then the initial velocity is positive and the final velocity is negative, so:

$\overrightarrow{J}=-(2kg)(10\frac{m}{s}) - (2.0kg)(10\frac{m}{s})$

$\overrightarrow{J}=-40 \frac{kg*m}{s}$

So, the impulse delivered to the floor is $40 \frac{kg*m}{s}$