The momentum versus time graph of object A originelly moving horizanty East is shown Wrete down the definition of momendum in words. 2 The net force acting on objed \( A \) is zero beturen \( L=10 \) s und \( t=20 \) s, use the graph and a relevant equation to explain why this statement is tiue. Calculate the megnitude of the impulse that

Certainly! Let's address each part of your question step by step. --- ## 1. Definition of Momentum in Words **Momentum** is a measure of an object's motion. It quantifies how much motion an object has and how difficult it would be to stop it. In more precise terms: **Momentum** is the product of an object's mass and its velocity. It is a vector quantity, meaning it has both magnitude and direction. Momentum depends directly on how much mass an object has and how fast it's moving in a particular direction. *Mathematically:* \[ \text{Momentum} (\mathbf{p}) = \text{mass} (m) \times \text{velocity} (\mathbf{v}) \] \[ \mathbf{p} = m \mathbf{v} \] --- ## 2. Net Force Acting on Object A Between \( t = 10 \) s and \( t = 20 \) s ### Understanding the Relationship Between Momentum and Force Newton's Second Law of Motion connects force and momentum. It states that the net force acting on an object is equal to the rate of change of its momentum with respect to time. *Mathematically:* \[ \mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} \] Where: - \( \mathbf{F}_{\text{net}} \) is the net force. - \( \mathbf{p} \) is the momentum. - \( t \) is time. ### Analyzing the Momentum vs. Time Graph - **Zero Net Force Implication:** If the net force acting on an object is zero over a certain time interval, the rate of change of its momentum during that interval is zero. This means the object's momentum remains constant during that time. - **Graph Interpretation:** On a **Momentum vs. Time graph**, a constant momentum is represented by a **horizontal line** (i.e., a line with zero slope) over the time interval of interest. - **Between \( t = 10 \) s and \( t = 20 \) s:** Since the net force is stated to be zero between these times, the momentum of object A does not change during this interval. Thus, on the graph, the momentum vs. time line should be horizontal between \( t = 10 \) s and \( t = 20 \) s. ### Conclusion The **horizontal line** (zero slope) on the momentum vs. time graph between \( t = 10 \) s and \( t = 20 \) s indicates that there is **no change in momentum** during this interval. According to Newton's Second Law, if the momentum is not changing, the **net force** acting on the object must be **zero**. Therefore, the statement that the net force acting on object A is zero between \( t = 10 \) s and \( t = 20 \) s is **true**. --- ## 3. Calculating the Magnitude of Impulse **Impulse** is defined as the change in momentum of an object when a force is applied over a period of time. It is given by the product of the average net force acting on the object and the time duration over which it acts. *Mathematically:* \[ \text{Impulse} (\mathbf{J}) = \Delta \mathbf{p} = \mathbf{F}_{\text{net}} \times \Delta t \] Where: - \( \Delta \mathbf{p} \) is the change in momentum. - \( \mathbf{F}_{\text{net}} \) is the net force. - \( \Delta t \) is the time interval. ### Given the Information - **Between \( t = 10 \) s and \( t = 20 \) s:** - Net force, \( \mathbf{F}_{\text{net}} = 0 \) N (from part 2). Since the net force is zero during this interval, the impulse exerted on the object during this time is: \[ \mathbf{J} = \mathbf{F}_{\text{net}} \times \Delta t = 0 \, \text{N} \times (20 \, \text{s} - 10 \, \text{s}) = 0 \, \text{N·s} \] ### Conclusion The **magnitude of the impulse** acting on object A between \( t = 10 \) s and \( t = 20 \) s is **zero** Newton-seconds (0 N·s) because there is **no net force** acting on the object during this interval, resulting in **no change in momentum**. --- **Note:** If you have specific numerical values from the momentum vs. time graph outside the \( t = 10 \) s to \( t = 20 \) s interval or additional information about forces acting outside this interval, you can calculate the impulse over those periods using the change in momentum. If there's more to the question or additional graphs/data, please provide them for a more detailed solution.