A body of mass $2$ kg moving with velocity of $\vec{v}_{in} = 3\hat{i} + 4\hat{j}$ ms$^{-1}$ enters into a constant force field of $6$ N 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) $3\hat{i} + 4\hat{j} + \sqrt{5}\hat{k}$
B) $4\hat{i} + 3\hat{j} + 5\hat{k}$
C) $3\hat{i} + 4\hat{j} – 5\hat{k}$
D) $3\hat{i} + 4\hat{j} + 5\hat{k}$
A) $3\hat{i} + 4\hat{j} + \sqrt{5}\hat{k}$
B) $4\hat{i} + 3\hat{j} + 5\hat{k}$
C) $3\hat{i} + 4\hat{j} – 5\hat{k}$
D) $3\hat{i} + 4\hat{j} + 5\hat{k}$
✓
Solution Steps
1
Define Initial Parameters
The initial velocity vector is given as:
$$\vec{u} = 3\hat{i} + 4\hat{j} \text{ m/s}$$
Mass of the body $m = 2 \text{ kg}$. The force $\vec{F}$ is $6 \text{ N}$ along the positive z-axis, which is written as:
$$\vec{F} = 6\hat{k} \text{ N}$$
2
Calculate Acceleration
Using Newton’s Second Law ($\vec{F} = m\vec{a}$), the acceleration vector $\vec{a}$ is:
$$\vec{a} = \frac{\vec{F}}{m} = \frac{6\hat{k}}{2} = 3\hat{k} \text{ m/s}^2$$
3
Identify Constant Components
Since the acceleration is only in the z-direction ($\hat{k}$), the velocity components in the x ($\hat{i}$) and y ($\hat{j}$) directions will remain constant during the motion.
4
Apply First Equation of Motion
The final velocity $\vec{v}$ after time $t = \frac{5}{3}$ s is given by $\vec{v} = \vec{u} + \vec{a}t$:
$$\vec{v} = (3\hat{i} + 4\hat{j}) + (3\hat{k}) \times \left(\frac{5}{3}\right)$$
5
Simplify the Expression
Perform the multiplication for the z-component:
$$\vec{v} = 3\hat{i} + 4\hat{j} + \left(3 \times \frac{5}{3}\right)\hat{k}$$
$$\vec{v} = 3\hat{i} + 4\hat{j} + 5\hat{k} \text{ m/s}$$
Final Result: The velocity is $3\hat{i} + 4\hat{j} + 5\hat{k}$.
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Theory
1. Motion in Three Dimensions
Motion in 3D can be complex, but it is effectively analyzed by decomposing the vectors into mutually perpendicular components (x, y, and z). According to the principle of superposition, the motion along each axis is independent of the others. In this problem, the object initially moves in the xy-plane. When it enters a force field acting exclusively along the z-axis, it gains a vertical velocity component while maintaining its horizontal velocity components. This independence allows us to use standard kinematic equations separately for each dimension to find the resultant vector.
2. Newton’s Second Law in Vector Form
Newton’s Second Law, $\vec{F} = m\vec{a}$, states that the acceleration produced in a body is directly proportional to the net force and inversely proportional to its mass. Crucially, the acceleration vector $\vec{a}$ always points in the same direction as the net force vector $\vec{F}$. If a force is applied along a specific unit vector (like $\hat{k}$), the body will only accelerate in that direction. This is why, in this scenario, the velocity components $v_x$ and $v_y$ remained unchanged, as there were no force components acting along the $\hat{i}$ or $\hat{j}$ directions.
3. Constant Force Fields
A constant force field is a region where every point exerts the same force in the same direction on an object. Common examples include the uniform gravitational field near Earth’s surface or a uniform electric field between two parallel plates. Because the force is constant, the resulting acceleration is also constant ($a = F/m$). This allows for the direct application of the equations of motion ($v = u + at$, etc.). In competitive exams like JEE, recognizing a constant force field is the key to simplifying the problem into a linear acceleration task.
4. Vector Addition and Kinematics
The final state of an object’s motion is determined by the vector sum of its initial state and the changes caused by external influences. The final velocity vector $\vec{v} = \vec{u} + \Delta\vec{v}$, where $\Delta\vec{v}$ is the change in velocity (impulse per unit mass). By calculating the change component-wise, we ensure that the directional integrity of the motion is preserved. Understanding that $\Delta\vec{v}$ is simply $\vec{a}t$ for constant acceleration helps in quickly finding the emerging velocity vector without needing to integrate complex functions.
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FAQs
1
Why did $3\hat{i}$ and $4\hat{j}$ not change?
Forces only change velocity in their own direction. Since the force was in the z-direction, x and y components remained constant.
2
What would happen if the force was directed along $-\hat{k}$?
The acceleration would be $-3\hat{k}$, and the final velocity would have been $3\hat{i} + 4\hat{j} – 5\hat{k}$.
3
Does the mass of the body matter?
Yes, mass determines the magnitude of acceleration for a given force ($a = F/m$). Higher mass leads to lower acceleration.
4
What is the magnitude of the final velocity?
It can be found using $|\vec{v}| = \sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50} = 5\sqrt{2}$ m/s.
5
Is the acceleration due to gravity included?
Unless specified, we only consider the “constant force field” mentioned in the problem description.
6
What if the force field was not constant?
We would have to use integration ($\int \vec{a} dt$) to find the change in velocity.
7
What is the unit of impulse in this problem?
Impulse $J = F \times t = 6 \times (5/3) = 10$ Ns.
8
Can this be solved using the work-energy theorem?
Work-energy theorem gives the final speed (magnitude), but not the individual vector components directly.
9
Does the path of the body matter?
In a constant force field, the change in velocity depends only on the time spent in the field, not the specific trajectory.
10
What is the displacement of the body along the z-axis?
Using $s_z = \frac{1}{2}a_z t^2$, $s_z = \frac{1}{2}(3)(5/3)^2 = \frac{1}{2}(3)(25/9) = 25/6$ meters.
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