Problem 15 - Master Resource Book in JEE Main Physics

Question

In a two dimensional motion of a particle, the particle moves from point \(A\), position vector \(\mathbf{r}_{1}\) to point \(B\), position vector \(\mathbf{r}_{2}\). If the magnitudes of these vectors are respectively, \(r_{1}=3\) and \(r_{2}=4\) and the angles they make with the \(x\)-axis are \(\theta_{1}=75^{\circ}\) and \(15^{\circ}\), respectively, then find the magnitude of the displacement vector (a) 15 (b) \(\sqrt{13}\) (c) 17 (d) \(\sqrt{15}\)

Step-by-Step Solution

Answer

The magnitude of the displacement vector is \(\sqrt{13}\), option (b).

Step 1: Calculate the components of vector \( \mathbf{r}_1 \)

The position vector \( \mathbf{r}_1 \), has a magnitude of \( r_1 = 3 \) and makes an angle of \( \theta_1 = 75^\circ \) with the x-axis. The x-component is \( r_{1x} = r_1 \cos \theta_1 = 3 \cos 75^\circ \) and the y-component is \( r_{1y} = r_1 \sin \theta_1 = 3 \sin 75^\circ \). Calculating these gives us:\[ r_{1x} = 3 \times 0.2588 \approx 0.776\] \[ r_{1y} = 3 \times 0.9659 \approx 2.897\]

Step 2: Calculate the components of vector \( \mathbf{r}_2 \)

The position vector \( \mathbf{r}_2 \), has a magnitude of \( r_2 = 4 \) and makes an angle of \( \theta_2 = 15^\circ \) with the x-axis. The x-component is \( r_{2x} = r_2 \cos \theta_2 = 4 \cos 15^\circ \) and the y-component is \( r_{2y} = r_2 \sin \theta_2 = 4 \sin 15^\circ \). Calculating these gives us:\[ r_{2x} = 4 \times 0.9659 \approx 3.864\] \[ r_{2y} = 4 \times 0.2588 \approx 1.035\]

Step 3: Calculate the components of the displacement vector \( \mathbf{d} \)

The displacement vector \( \mathbf{d} \) from \( A \) to \( B \) is given by \( \mathbf{d} = \mathbf{r}_2 - \mathbf{r}_1 \). Therefore, its components are:\[ d_x = r_{2x} - r_{1x} = 3.864 - 0.776 = 3.088 \]\[ d_y = r_{2y} - r_{1y} = 1.035 - 2.897 = -1.862 \]

Step 4: Calculate the magnitude of the displacement vector \( \mathbf{d} \)

The magnitude of the displacement vector \( \mathbf{d} \) is given by the Pythagorean theorem:\[ |\mathbf{d}| = \sqrt{d_x^2 + d_y^2} = \sqrt{3.088^2 + (-1.862)^2} \]Calculate:\[ d_x^2 = 3.088^2 \approx 9.535 \]\[ d_y^2 = (-1.862)^2 \approx 3.468 \]Thus, \[ |\mathbf{d}| = \sqrt{9.535 + 3.468} = \sqrt{13.003} \approx \sqrt{13} \]

Key Concepts

Two-Dimensional Motion

In two-dimensional motion, an object moves in a plane. This means that its position changes in two dimensions—typically described using the x-axis and y-axis. Such motion is common in many real-world scenarios, from sports to car driving, where the movement is not confined to a single straight line. Each position of the moving object can be represented by a pair of coordinates \(x, y\). The advantage of visualizing motion in two dimensions is the ability to consider separate influences along each axis, like gravity or other forces.
By breaking down the motion into two components, you can understand how an object is moving both horizontally and vertically at the same time. This helps in predicting future positions and understanding the nature of the object's trajectory, making two-dimensional motion a fundamental part of physics studies.

Position Vectors

Position vectors are used to define the position of a point in space relative to an origin. In the context of two-dimensional motion, a position vector is represented by the components along the x and y axes.
For instance, in this problem, the particle moves from point A to point B. Each position is represented by vectors \(\mathbf{r}_1\) and \(\mathbf{r}_2\). These vectors are determined by their magnitudes and the angles they make with a reference axis, usually the x-axis. Magnitude gives the length of the vector, while the angle determines the vector's direction.
Understanding position vectors helps in determining where exactly an object is situated in the space and how it travels from one point to another. This is crucial in analyzing motions and related areas like navigation and flight dynamics.

Angle Calculation

Angles are essential in determining the orientation of vectors in two-dimensional space. In the exercise, the angles \(\theta_1\) and \(\theta_2\) define the position vectors' directions from the x-axis. \(\theta_1 = 75^\circ\) and \(\theta_2 = 15^\circ\).
Calculating the vector's components involves using trigonometric functions—cosine and sine—linked to these angles. The cosine function helps find the horizontal component, while the sine function finds the vertical component. For example, for a vector with magnitude \(r_1\) and an angle \(\theta_1\), we find the components as follows:

• \(r_{1x} = r_1 \cos(\theta_1)\)
• \(r_{1y} = r_1 \sin(\theta_1)\)

Angle calculations ensure that the vectors are represented accurately, which is vital for determining effects like direction and interactions with other forces or vectors.

Vector Components

Every vector in a two-dimensional space can be broken down into its components along the x and y axes, simplifying calculations and understanding the motion.
In this case, to find the displacement vector from \(\mathbf{r}_1\) to \(\mathbf{r}_2\), it's essential first to find the components of each position vector. By using trigonometric identities, we calculate these as follows:

• The x-component: \(r_{1x} = r_1 \cos(75^\circ)\) and \(r_{2x} = r_2 \cos(15^\circ)\)
• The y-component: \(r_{1y} = r_1 \sin(75^\circ)\) and \(r_{2y} = r_2 \sin(15^\circ)\)

Once the components are found, the displacement vector \(\mathbf{d}\) can be computed by subtracting corresponding components:

• Horizontal displacement: \(d_x = r_{2x} - r_{1x}\)
• Vertical displacement: \(d_y = r_{2y} - r_{1y}\)

Understanding vector components is key to solving various physics problems, as it allows manipulating vectors in an analytical way, essential for predicting motion and understanding force interactions.