Problem 41 - Master Resource Book in JEE Main Physics

Question

A proton in a cyclotron changes its velocity from \(30 \mathrm{kms}^{-1}\) due north to \(40 \mathrm{kms}^{-1}\) due east in \(20 \mathrm{~s}\). What is the magnitude of average acceleration during this time? (a) \(2.5 \mathrm{kms}^{-2}\) (b) \(12.5 \mathrm{kms}^{-2}\) (c) \(22.5 \mathrm{kms}^{-2}\) (d) \(32.5 \mathrm{kms}^{-2}\)

Step-by-Step Solution

Answer

The correct answer is (a) \(2.5 \, \mathrm{kms}^{-2}\).

Step 1: Identify Initial and Final Velocities

The initial velocity of the proton is given as \(30 \ \mathrm{kms}^{-1}\) due north, which we can represent as a vector \(\mathbf{v}_i = 30 \ \hat{j} \ \mathrm{kms}^{-1}\). The final velocity is \(40 \ \mathrm{kms}^{-1}\) due east, represented as \(\mathbf{v}_f = 40 \ \hat{i} \ \mathrm{kms}^{-1}\).

Step 2: Calculate Change in Velocity

The change in velocity \(\Delta \mathbf{v}\) is the difference between the final and initial velocities: \[\Delta \mathbf{v} = \mathbf{v}_f - \mathbf{v}_i = 40 \ \hat{i} - 30 \ \hat{j} \ \mathrm{kms}^{-1}.\]This results in a vector change of \(\Delta \mathbf{v} = 40 \ \hat{i} - 30 \ \hat{j} \ \mathrm{kms}^{-1}\).

Step 3: Calculate Magnitude of Change in Velocity

Use the Pythagorean theorem to find the magnitude of the change in velocity: \[\|\Delta \mathbf{v}\| = \sqrt{(40)^2 + (-30)^2} = \sqrt{1600 + 900} = \sqrt{2500} = 50 \ \mathrm{kms}^{-1}.\]

Step 4: Calculate Average Acceleration

Average acceleration \(\mathbf{a}_{avg}\) is given by the change in velocity divided by the time interval: \[\mathbf{a}_{avg} = \frac{\|\Delta \mathbf{v}\|}{\Delta t} = \frac{50 \ \mathrm{kms}^{-1}}{20 \, \mathrm{s}} = 2.5 \, \mathrm{kms}^{-2}.\]

Step 5: Select Correct Answer from Options

The magnitude of average acceleration calculated is \(2.5 \, \mathrm{kms}^{-2}\), which matches option (a).

Key Concepts

Cyclotron

Imagine a large, circular device where charged particles, like protons, speed around and gain energy. This fascinating machine is known as a cyclotron. Cyclotrons use a magnetic field to keep the charged particles on a spiral path. Alongside, an electric field is used to accelerate these particles to higher speeds.

In simpler terms, think of it as a giant merry-go-round where protons are pushed faster and faster each time they make a complete circle. The combination of magnetic and electric fields makes this possible.

• The magnetic field bends the path of the proton, keeping it in a spiral trajectory.
• The electric field accelerates the proton, increasing its velocity with each turn.

This mechanism is widely used in scientific research for speeding up particles to very high speeds, which is necessary for experiments in nuclear physics and medicine.

Velocity Change

Velocity is more than just speed; it is speed with a direction. When we talk about a change in velocity, it means there has been a change in either the magnitude or direction of speed, or both. Consider our initial example with the proton changing its speed and direction - from due north to due east.

The difference between these two velocities gives us the velocity change, which can be represented as a vector. This vector points from the initial velocity direction to the final velocity direction.

• Initial velocity: 30 km/s north (\(30 \hat{j}\ km/s\)).
• Final velocity: 40 km/s east (\(40 \hat{i}\ km/s\)).
• The change in velocity is calculated by subtracting the two vectors: \(\Delta \mathbf{v} = 40 \hat{i} - 30 \hat{j} \ km/s\).

The magnitude of this change gives us an idea of how much the velocity has altered. In this case, the magnitude is 50 km/s as calculated using the Pythagorean theorem.

Motion in Two Dimensions

When an object moves in more than one direction, it is said to be in motion in two dimensions. This is a common scenario, as most real-world movements aren't limited to a single straight line. In the case of our proton in the cyclotron, its movement can be described in terms of two dimensions: north-south and east-west.

Handling motion in two dimensions often requires understanding and dealing with vectors. Vectors are essential because they consider both the magnitude and direction, capturing the essence of multi-directional motion.

• Each direction can be treated as a separate component (for example, north-south as one and east-west as another).
• It's like observing two independent motions at the same time, which together describe the complete path.
• Mathematical techniques like vector addition are crucial in analyzing and solving two-dimensional motion problems.

Understanding two-dimensional motion is vital in physics as it applies to everything from sports trajectories to the paths charged particles take in fields.