Question

A body of mass 2 kg moving with velocity of $ \vec{v}_{\text{in}} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1} $ enters into a constant force field of 6N directed along positive z-axis. If the body remains in the field for a period of $ \frac{5}{3} $ seconds, then velocity of the body when it emerges from force field is:

• A. \( 3 \hat{i} + 4 \hat{j} - 5 \hat{k} \)
• B. \( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \)
• C. \( 3 \hat{i} + 4 \hat{j} + \sqrt{5} \hat{k} \)
• D. \( 4 \hat{i} + 3 \hat{j} + 5 \hat{k} \)

Hint:

In problems involving constant forces, use Newton's second law to find acceleration and apply the equations of motion to find the final velocity.

Answer 1

The Correct Option is B

To solve this question, we need to find the velocity of a body after it has been influenced by a constant force for a specific duration.

Given:

• Initial velocity of the body: \(\vec{v}_{\text{in}} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1}\)
• Mass of the body: \(m = 2 \, \text{kg}\)
• Force acting on the body: \(\vec{F} = 6 \hat{k} \, \text{N}\)
• Time duration in the force field: \(t = \frac{5}{3} \, \text{s}\)

To find the final velocity, we can use the equation of motion under constant acceleration:

\(\vec{v}_{\text{final}} = \vec{v}_{\text{in}} + \vec{a} \cdot t\)

First, let's calculate the acceleration using Newton's second law:

\(\vec{F} = m \cdot \vec{a} \Longrightarrow \vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \, \text{ms}^{-2}\)

Substitute the values into the velocity equation:

\(\vec{v}_{\text{final}} = (3 \hat{i} + 4 \hat{j}) + (3 \hat{k}) \cdot \frac{5}{3}\)

Further simplify the equation:

\(\vec{v}_{\text{final}} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k}\)

Thus, the velocity of the body when it emerges from the force field is: \(3 \hat{i} + 4 \hat{j} + 5 \hat{k}\).

Hence, the correct answer is:

\( 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \)

 

Answer 2

The force on the body is given as \( \vec{F} = 6 \hat{k} \, \text{N} \), which is directed along the positive z-axis. Using Newton's second law \( \vec{F} = m \vec{a} \), where \( m \) is the mass of the body: \[ \vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} \, \text{ms}^{-2} \] Thus, the acceleration of the body is \( 3 \hat{k} \, \text{ms}^{-2} \), which means the body accelerates in the positive z-direction. To find the final velocity, we use the equation of motion: \[ \vec{v}_{\text{final}} = \vec{v}_{\text{initial}} + \vec{a} \Delta t \] Substituting the given values: \[ \vec{v}_{\text{initial}} = 3 \hat{i} + 4 \hat{j} \, \text{ms}^{-1}, \quad \vec{a} = 3 \hat{k} \, \text{ms}^{-2}, \quad \Delta t = \frac{5}{3} \, \text{seconds} \] \[ \vec{v}_{\text{final}} = (3 \hat{i} + 4 \hat{j}) + (3 \hat{k}) \times \frac{5}{3} \] \[ \vec{v}_{\text{final}} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} \] Thus, the final velocity of the body when it emerges from the force field is: \[ \boxed{3 \hat{i} + 4 \hat{j} + 5 \hat{k}} \]