8. To calculate the net impulse that acted on the ball, we need to use the concept of impulse and momentum. Impulse (J) can be defined as the change in momentum of an object when a force is applied over a period of time. It can be calculated using the following formula:
\[ J = \Delta p \]
where \(\Delta p\) is the change in momentum of the object. Momentum (p) is given by the product of the mass (m) of the object and its velocity (v), i.e.:
\[ p = m \times v \]
In this case, the ball has a mass of 100 kg, an initial velocity (\(v_i\)) of 10 m/s, and a final velocity (\(v_f\)) of 15 m/s. The change in velocity (\(\Delta v\)) is:
\[ \Delta v = v_f - v_i = 15\, \text{m/s} - 10\, \text{m/s} = 5\, \text{m/s} \]
The change in momentum (\(\Delta p\)) is then:
\[ \Delta p = m \times \Delta v = 100\, \text{kg} \times 5\, \text{m/s} = 500\, \text{kg}\cdot\text{m/s} \]
Therefore, the net impulse (J) that acted on the ball is:
\[ J = \Delta p = 500\, \text{kg} \cdot \text{m/s} \]
9. To find the impulse delivered to a system by an external force, we use the following formula for impulse (J), which relates force (F) applied over a time interval (t):
\[ J = F \times t \]
Given that an external force of \(F = 34.0\, \text{N}\) acts on the system for \(t = 12.0\, \text{s}\), we can calculate the impulse as:
\[ J = 34.0\, \text{N} \times 12.0\, \text{s} = 408.0\, \text{N}\cdot\text{s} \]
Since the force acts in the negative x direction, the impulse would also be directed in the negative x direction. Therefore, if we consider the direction, the impulse delivered to the system is:
\[ J = -408.0\, \text{N}\cdot\text{s} \]
Please note that the negative sign indicates direction and does not affect the magnitude of the impulse.
