Question

A particle of mass $0.5\text{ kg}$ undergoes a collision where its velocity changes from $4\hat{i}\text{ m/s}$ to $-3\hat{j}\text{ m/s}$. The magnitude of impulse imparted to the particle is:

• A. 2.5 N s
• B. 3.5 N s
• C. 1.5 N s
• D. 5.0 N s Correct Answer: (A) 2.5 N s

Hint:

Whenever you see a vector velocity shift from a pure horizontal component ($4\hat{i}$) to a pure vertical component ($-3\hat{j}$), look for the classic 3-4-5 right triangle pattern! The net magnitude of the change is instantly $5\text{ m/s}$. Multiplying this total velocity swing by a mass of $0.5\text{ kg}$ (which means dividing by 2) yields 2.5 N s within seconds!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
When an object experiences a sudden force, such as during a collision, it undergoes a rapid change in its motion. In classical mechanics, this event is described by the Impulse-Momentum Theorem. The total impulse ($\vec{J}$) delivered to a particle equals its net change in linear momentum ($\Delta \vec{p}$). Because velocity and momentum are vector quantities possessing both directional orientation and numeric value, tracking a multi-dimensional change requires proper vector subtraction rather than basic scalar math.

Step 2: Key Formula or Approach:
The formulas governing vector impulse and momentum are: 1. Linear Momentum ($\vec{p}$): $$ \vec{p} = m\vec{v} $$ 2. Impulse-Momentum Theorem: $$ \vec{J} = \Delta \vec{p} = \vec{p}_f - \vec{p}_i = m(\vec{v}_f - \vec{v}_i) $$ 3. Vector Magnitude: For any vector $\vec{A} = A_x\hat{i} + A_y\hat{j}$, its absolute magnitude is: $$ |\vec{A}| = \sqrt{A_x^2 + A_y^2} $$ Let's extract the variables provided: - Mass of the particle ($m$) = $0.5\text{ kg}$ - Initial velocity vector ($\vec{v}_i$) = $4\hat{i}\text{ m/s}$ - Final velocity vector ($\vec{v}_f$) = $-3\hat{j}\text{ m/s}$

Step 3: Detailed Explanation:
Let's construct and solve the vector equation systematically: 1. Calculate the change in velocity vector ($\Delta \vec{v}$): $$ \Delta \vec{v} = \vec{v}_f - \vec{v}_i = (-3\hat{j}) - (4\hat{i}) = -4\hat{i} - 3\hat{j}\text{ m/s} $$ 2. Calculate the absolute magnitude of this change in velocity ($|\Delta \vec{v}|$): Using the Pythagorean theorem for perpendicular vector components: $$ |\Delta \vec{v}| = \sqrt{(-4)^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\text{ m/s} $$ 3. Calculate the magnitude of the imparted impulse ($|\vec{J}|$): Multiply the mass by the calculated velocity change magnitude: $$ |\vec{J}| = m \times |\Delta \vec{v}| = 0.5\text{ kg} \times 5\text{ m/s} = 2.5\text{ N s} $$ The evaluated impulse magnitude is exactly $2.5\text{ N s}$, matching option (A).

Step 4: Final Answer:
The magnitude of the impulse imparted to the particle is $2.5\text{ N s}$.