Physics Q27 – JEE-MAIN 2025

Home Physics Laws of MotionQuestion #229

A body of mass 2 kg moving with velocity

{\vec{v}}_{i n} = 3 \hat{i} + 4 \hat{j} {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.

Clue 1 of 3: Vector Nature of Motion

Remember that force and velocity are vector quantities, and their components can be analyzed independently.

🥷

Ninja Strategy•Component Analysis

Recognize that a force acting purely in one direction only affects the velocity component in that direction, leaving perpendicular components unchanged.

🧠 Full Solution Path

Step 1: Calculate acceleration due to the force.✦ Active

The force $\vec{F} = 6 \hat{k}$ N acts on a mass $m = 2$ kg. Using Newton's second law, $\vec{F} = m \vec{a}$ , the acceleration is:

\vec{a} = \frac{\vec{F}}{m} = \frac{6 \hat{k}}{2} = 3 \hat{k} {ms}^{- 2}

Step 2: Calculate the change in velocity.○ Expand

The body remains in the field for $Δ t = \frac{5}{3}$ s. The change in velocity is given by $Δ \vec{v} = \vec{a} Δ t$ :

Δ \vec{v} = (3 \hat{k}) (\frac{5}{3}) = 5 \hat{k} {ms}^{- 1}

Step 3: Determine the final velocity.○ Expand

The initial velocity is ${\vec{v}}_{i n} = 3 \hat{i} + 4 \hat{j}$ ms $^{- 1}$ . The final velocity is the sum of the initial velocity and the change in velocity:

{\vec{v}}_{o u t} = {\vec{v}}_{i n} + Δ \vec{v} = (3 \hat{i} + 4 \hat{j}) + 5 \hat{k} = 3 \hat{i} + 4 \hat{j} + 5 \hat{k} {ms}^{- 1}

💡 Teacher's Secret Hint

Remember that components of velocity perpendicular to the force remain unchanged.

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Physics Question 27 – JEE-MAIN 2025

Clue 1 of 3: Vector Nature of Motion

🧠 Full Solution Path