Question

An object of mass \(3 \mathrm{~kg}\) is moving with a velocity of $5 \mathrm{~m} / \mathrm{s}\( along a straight path. If a force of \)12 \mathrm{~N}$ is applied for \(3 \mathrm{sec}\) on the object in a perpendicular to its direction of motion. The magnitude of velocity of the particle at the end of $3 \mathrm{sec}\( is \)\mathrm{m} / \mathrm{s}$. (A) (B) 12 (C) 13 (D) 4

Short answer

The magnitude of the final velocity of the object after the force is applied for 3 seconds is \(13 \: \mathrm{m/s}\), which corresponds to option (C).

Step-by-step solution

Identify given values

Here's a summary of the given information: - Mass of object (m): 3 kg - Initial velocity (v1): 5 m/s (along a straight path) - Force applied (F): 12 N (perpendicular to the direction of motion) - Time force is applied (t): 3 s

Determine acceleration due to perpendicular force

Using Newton's second law, we can express the acceleration due to the force (a2) as follows: \[a_2 = \frac{F}{m}\] where F = 12 N (force applied) m = 3 kg (mass of the object) Now we can find the acceleration: \[a_2 = \frac{12}{3} = 4 \: \mathrm{m/s^2}\]

Calculate the velocity resulting from the perpendicular force

We can use the equation for calculating final velocity (v2) after a given time (t) and acceleration (a2): \[v_2 = a_2 \times t \] Substituting the given values: \[v_2 = 4 \: \mathrm{m/s^2} \times 3 \: \mathrm{s} = 12 \: \mathrm{m/s}\]

Combine the initial velocity and the velocity due to the force

Since the initial velocity (v1) and the velocity due to the force (v2) are perpendicular to each other, we can use the Pythagorean theorem to find the magnitude of the final velocity (v): \[v = \sqrt{v_1^2 + v_2^2}\] Substituting the given values: \[v = \sqrt{(5 \: \mathrm{m/s})^2 + (12 \: \mathrm{m/s})^2} = \sqrt{169} = 13 \: \mathrm{m/s}\] The magnitude of the final velocity of the object after the force is applied for 3 seconds is 13 m/s, which corresponds to option (C).