Question

1. how can you express impulse in terms of mass and velocity when neither of those are constant?

a. $\delta \mathbf{p} = \delta (mv)$
b. $\frac{\delta \mathbf{p}}{\delta t} = \frac{\delta (mv)}{\delta t}$
c. $\delta \mathbf{p} = \delta (\frac{m}{v})$
d. $\frac{\delta \mathbf{p}}{\delta t} = \frac{1}{\delta t} \cdot \delta (mv)$

1. how can you express impulse in terms of mass and

Explanation:

Step1: Recall Impulse-Momentum Theorem

Impulse (\(J\)) is equal to the change in momentum (\(\Delta \mathbf{p}\)). Momentum \(\mathbf{p}\) is defined as \(m\mathbf{v}\) (mass times velocity). So, the change in momentum \(\Delta \mathbf{p}\) should be the change in \(m\mathbf{v}\), i.e., \(\Delta \mathbf{p}=\Delta(m\mathbf{v})\), regardless of whether \(m\) or \(\mathbf{v}\) is constant.

Step2: Analyze Each Option

• Option a: \(\Delta \mathbf{p}=\Delta(m\mathbf{v})\) matches the impulse - momentum theorem (impulse equals change in momentum, and momentum is \(m\mathbf{v}\)).
• Option b: \(\frac{\Delta \mathbf{p}}{\Delta t}=\frac{\Delta(m\mathbf{v})}{\Delta t}\) relates the rate of change of momentum (which is force, \(F = \frac{\Delta \mathbf{p}}{\Delta t}\)) to the rate of change of \(m\mathbf{v}\), not the expression for impulse in terms of \(m\) and \(\mathbf{v}\).
• Option c: \(\Delta \mathbf{p}=\Delta(\frac{m}{\mathbf{v}})\) is incorrect because momentum is \(m\mathbf{v}\), not \(\frac{m}{\mathbf{v}}\).
• Option d: \(\frac{\Delta \mathbf{p}}{\Delta t}=\frac{1}{\Delta t}\cdot\Delta(m\mathbf{v})\) is a mis - formed equation. The left - hand side is force (rate of change of momentum), and the right - hand side does not represent a valid physical relationship for impulse.

Answer:

a. \(\Delta \mathbf{p}=\Delta(m\mathbf{v})\)