Question

1. how can you express impulse in terms of mass and velocity when neither of those are constant? a. $\delta \mathbf{p}=\delta(m \mathbf{v})$ b. $\frac{\delta \mathbf{p}}{\delta t}=\frac{\delta(m \mathbf{v})}{\delta t}$ c. $\delta \mathbf{p}=\delta(\frac{m}{\mathbf{v}})$ d. $\frac{\delta \mathbf{p}}{\delta t}=\frac{1}{\delta t} \cdot \delta(m \mathbf{v})$ 35. how can you express impulse in terms of mass and initial and final velocities? a. $\delta \mathbf{p}=m(\mathbf{v}_{f}-\mathbf{v}_{i})$ b. $\frac{\delta \mathbf{p}}{\delta t}=\frac{m(\mathbf{v}_{f}-\mathbf{v}_{i})}{\delta t}$ c. $\delta \mathbf{p}=\frac{(\mathbf{v}_{f}-\mathbf{v}_{i})}{m}$ d. $\frac{\delta \mathbf{p}}{\delta t}=\frac{1}{m} \frac{(\mathbf{v}_{f}-\mathbf{v}_{i})}{\delta t}$

Explanation:

Step1: Recall the definition of impulse

Impulse $\Delta\mathbf{p}$ is defined as the change in momentum. Momentum is given by $\mathbf{p} = m\mathbf{v}$. So, when we want to find the change in momentum (impulse), we consider $\Delta\mathbf{p}=\mathbf{p}_{final}-\mathbf{p}_{initial}$. If $\mathbf{p}=m\mathbf{v}$, then $\Delta\mathbf{p}=\Delta(m\mathbf{v})$

Step2: Analyze each option for question 34

• Option a: $\Delta\mathbf{p}=\Delta(m\mathbf{v})$ is correct as per the definition of impulse (change in momentum)
• Option b: $\frac{\Delta\mathbf{p}}{\Delta t}=\frac{\Delta(m\mathbf{v})}{\Delta t}$ gives the force (since $F = \frac{\Delta\mathbf{p}}{\Delta t}$), not impulse
• Option c: $\Delta\mathbf{p}=\Delta(\frac{m}{\mathbf{v}})$ is not in line with the momentum formula $\mathbf{p}=m\mathbf{v}$
• Option d: $\frac{\Delta\mathbf{p}}{\Delta t}=\frac{1}{\Delta t}\cdot\Delta(m\mathbf{v})$ simplifies to $\frac{\Delta\mathbf{p}}{\Delta t}=\frac{\Delta(m\mathbf{v})}{\Delta t}$ (force formula)

Step3: Analyze each option for question 35

• Option a: $\Delta\mathbf{p}=m(\mathbf{v}_f - \mathbf{v}_i)$ is correct. Since $\Delta\mathbf{p}=\mathbf{p}_f-\mathbf{p}_i$ and $\mathbf{p}=m\mathbf{v}$, then $\Delta\mathbf{p}=m\mathbf{v}_f - m\mathbf{v}_i=m(\mathbf{v}_f - \mathbf{v}_i)$
• Option b: $\frac{\Delta\mathbf{p}}{\Delta t}=\frac{m(\mathbf{v}_f-\mathbf{v}_i)}{\Delta t}$ gives the force (using $F=\frac{\Delta\mathbf{p}}{\Delta t}$)
• Option c: $\Delta\mathbf{p}=\frac{(\mathbf{v}_f - \mathbf{v}_i)}{m}$ is not correct as per the momentum formula $\mathbf{p}=m\mathbf{v}$
• Option d: $\frac{\Delta\mathbf{p}}{\Delta t}=\frac{1}{m}\frac{(\mathbf{v}_f - \mathbf{v}_i)}{\Delta t}$ is also not in line with the impulse - momentum relation

Answer:

1. a. $\Delta\mathbf{p}=\Delta(m\mathbf{v})$
2. a. $\Delta\mathbf{p}=m(\mathbf{v}_f - \mathbf{v}_i)$