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: Find the start and end momentum.
Impulse is just the change in momentum. Momentum is mass times velocity. With $m=0.5\text{ kg}$, the start momentum is $0.5\times 4\hat{i} = 2\hat{i}$ and the end momentum is $0.5\times(-3\hat{j}) = -1.5\hat{j}$ (units N s).

Step 2: Subtract to get the impulse vector.
\[ \vec{J} = \vec{p}_f - \vec{p}_i = (-1.5\hat{j}) - (2\hat{i}) = -2\hat{i} - 1.5\hat{j} \]

Step 3: Take the size of this vector.
The two parts are at right angles, so use the right triangle rule: \[ |\vec{J}| = \sqrt{(-2)^2 + (-1.5)^2} = \sqrt{4 + 2.25} = \sqrt{6.25} \]

Step 4: Final value.
\[ |\vec{J}| = 2.5\text{ N s} \]
So the impulse given to the particle is 2.5 N s, which is option (A).
\[ \boxed{2.5\text{ N s}} \]