Problem 25 - Master Resource Book in JEE Main Physics

Question

A particle is projected with a velocity of \(30 \mathrm{~m} / \mathrm{s}\), at an angle \(\theta_{0}=\tan ^{-1}\left(\frac{3}{4}\right)\). After \(1 \mathrm{~s}\), the particle is moving at an angle \(\theta\) to the horizontal, where \(\tan \theta\) will be equal to \(\left(g=10 \mathrm{~m} / \mathrm{s}^{2}\right)\) (a) \(]\) (b) 2 (c) \(\frac{1}{2}\) (d) \(\frac{1}{3}\)

Step-by-Step Solution

Answer

The answer is (d) \(\frac{1}{3}\).

Step 1: Identify Initial Velocity Components

The initial speed of the particle is given as \(30 \, \text{m/s}\). We need to find the horizontal (\(u_x\)) and vertical (\(u_y\)) components of this velocity. Given the angle \(\theta_0 = \tan^{-1}\left(\frac{3}{4}\right)\), we have \(\tan \theta_0 = \frac{3}{4}\). This gives us the vertical-to-horizontal component ratio. Using the identity, \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}}\), the values are \(u_y = \frac{3}{5} \times 30\) and \(u_x = \frac{4}{5} \times 30\).

Step 2: Calculate Initial Velocity Components

Calculate: \(u_y = \frac{3}{5} \times 30 = 18 \, \text{m/s} \) and \(u_x = \frac{4}{5} \times 30 = 24 \, \text{m/s} \). These are the initial velocity components.

Step 3: Update Velocity Components After 1 Second

After 1 second, the horizontal component of velocity remains \(u_x = 24 \, \text{m/s}\) because horizontal velocity is unaffected by gravity. The vertical component becomes \(v_y = u_y - gt\), where \(g = 10 \, \text{m/s}^2\) and \(t = 1\). Thus, \(v_y = 18 - 10 \times 1 = 8 \, \text{m/s}\).

Step 4: Find the Tangent of Resultant Angle

We need to find \(\tan \theta\) of the velocity vector after 1 second. This is \(\tan \theta = \frac{v_y}{u_x} = \frac{8}{24} = \frac{1}{3}\).

Step 5: Select the Correct Option

When comparing \(\tan \theta = \frac{1}{3}\) to the given choices, (a) does not specify a value, (b) is 2, (c) is \(\frac{1}{2}\), and (d) is \(\frac{1}{3}\). Thus, the correct answer is option (d).

Step 6: Verify Calculation

Check calculations step-by-step to confirm no arithmetic errors have been made in defining \(u_y, u_x\), and \(v_y\) to ensure that \(\tan \theta = \frac{1}{3}\). This matches with option (d).

Key Concepts

Initial Velocity Components

When a particle is launched with a certain velocity, it is helpful to break down this velocity into two perpendicular components: horizontal and vertical. These are referred to as the initial velocity components. For a projectile motion problem, this is especially crucial since these components respond differently to external forces like gravity.

• Horizontal Component (\(u_x\)): This indicates how fast the particle is moving horizontally. It remains constant during projectile motion (ignoring air resistance) because gravity does not act horizontally.
• Vertical Component (\(u_y\)): This shows how fast the particle is moving vertically. It changes over time as gravity acts downward, affecting the particle’s movement.

Given an angle of projection and the initial velocity, trigonometric functions can help determine these components.
For example, if an initial speed is \(30 ext{ m/s}\) and the angle \(\theta_0 = \tan^{-1}\left(\frac{3}{4}\right)\), you can calculate:

• \(u_x = 30 \times \cos \theta_0 = 24 \text{ m/s}\)
• \(u_y = 30 \times \sin \theta_0 = 18 \text{ m/s}\)

Motion Under Gravity

Motion under gravity refers to how an object moves when it is only subjected to gravitational pull, neglecting air resistance. In projectile motion, gravity plays a significant role in altering the trajectory of the particle by affecting its vertical motion.
Gravity is a constant force that acts downward with an acceleration of \(g = 9.8 \text{ m/s}^2\) (commonly approximated as \(10 \text{ m/s}^2\) in simpler calculations). Here's how it affects a projectile:

• The horizontal motion (\(u_x\)) stays constant because gravity does not act in the horizontal plane.
• The vertical motion (\(u_y\)) decreases as the particle ascends, eventually reaching zero at the peak of the trajectory, and then increases in the negative direction during descent.

After 1 second, the vertical velocity component \(v_y\) can be calculated using the formula:\[v_y = u_y - g \times t\]For example, if \(u_y = 18 \text{ m/s}\) and \(t = 1 \text{ s}\), then:\[v_y = 18 - 10 = 8 \text{ m/s}\]This illustrates that gravity reduces the vertical component as time progresses.

Kinematics

Kinematics is the branch of physics that deals with the motion of objects without considering the forces that cause them to move. It is crucial in analyzing projectile motion, particularly in predicting position and velocity over time.
In kinematics, equations of motion are used to anticipate how an object's velocity and position evolve, given initial conditions. With projectile motion, elements of kinematics help us to:

• Calculate initial and final velocities
• Determine the displacement over time
• Evaluate the effect of acceleration due to gravity

For instance, the vertical motion of a projectile can be examined using the equation:\[v = u + at\]Where:- \(v\) is the final vertical velocity- \(u\) is the initial vertical velocity- \(a\) is the acceleration (which can be gravity)- \(t\) is timeThis allows us to determine how fast the projectile is moving upward or downward after a given time.

Trigonometric Functions

Trigonometric functions are mathematical tools that relate the angles of a triangle to the lengths of its sides. In the context of projectile motion, they are essential for computing the initial velocity components of a projectile.
Three primary functions are often used:

• **Sine (\(\sin\)):** Used to find the vertical component of the velocity.
• **Cosine (\(\cos\)):** Used to find the horizontal component of the velocity.
• **Tangent (\(\tan\)):** Used to find the ratio between the vertical and horizontal components.

For a projectile launched at an angle \(\theta_0\) with velocity \(v_0\):

• The horizontal component is \(v_0 \cos \theta_0\)
• The vertical component is \(v_0 \sin \theta_0\)

In our problem, the initial projection angle is given by \(\tan^{-1}\left(\frac{3}{4}\right)\). This provides the ratio \(\frac{3}{4}\), which helps to compute both velocity components precisely. Thus, trigonometric functions are vital for understanding the initial setup of a projectile’s path.