Solution: Work Done by Variable Impulse

1. Physics Principles

The Work-Energy Theorem states that the total work done on an object is equal to the change in its kinetic energy:

$$ W_{net} = \Delta K = K_{final} – K_{initial} $$

We can find the final velocity using the Impulse-Momentum relationship: $\vec{p}_f = \vec{p}_i + \vec{J}$.

2. Visual Analysis

3. Calculation

Initial State:

• Mass $m = 2.0$ kg
• Velocity $\vec{v}_i = 5.0 \hat{i}$ m/s (East)
• Initial Momentum $\vec{p}_i = m\vec{v}_i = 10 \hat{i}$ Ns

Impulse Applied:

• $\vec{J} = -18.0 \hat{i} + 6.0 \hat{j}$ Ns (West is $-\hat{i}$, North is $+\hat{j}$)

Final State:

$$ \vec{p}_f = \vec{p}_i + \vec{J} $$ $$ \vec{p}_f = (10 \hat{i}) + (-18 \hat{i} + 6 \hat{j}) $$ $$ \vec{p}_f = -8 \hat{i} + 6 \hat{j} \text{ Ns} $$

Magnitude of final momentum:

$$ |\vec{p}_f| = \sqrt{(-8)^2 + (6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \text{ Ns} $$

Kinetic Energy Calculation:

$$ K_i = \frac{p_i^2}{2m} = \frac{10^2}{2(2)} = \frac{100}{4} = 25 \text{ J} $$ $$ K_f = \frac{p_f^2}{2m} = \frac{10^2}{2(2)} = \frac{100}{4} = 25 \text{ J} $$

Work Done:

$$ W = \Delta K = K_f – K_i = 25 – 25 = 0 \text{ J} $$