Question:

A 0.2 kg object is initially at rest when it is struck by an impulse. The impulse causes the object to move with a velocity of 3 m/s in the positive x-direction. The impulse acts for a duration of 0.1 seconds. Calculate the magnitude and direction of the impulse.

Answer:

Given: Mass of the object, m = 0.2 kg Initial velocity, u = 0 m/s Final velocity, v = 3 m/s Time, Δt = 0.1 s

To find the magnitude of impulse, we will use the impulse-momentum relation:

Impulse (J) = Force (F) x Δt = mΔv

We can calculate the change in velocity (Δv) using the equation: Δv = v - u

Substituting the given values: Δv = 3 m/s - 0 m/s Δv = 3 m/s

Now, we can calculate the magnitude of the impulse using the relation: J = mΔv

Substituting the values: J = (0.2 kg)(3 m/s) J = 0.6 kg·m/s

Therefore, the magnitude of the impulse is 0.6 kg·m/s.

To find the direction of the impulse, we need to consider the positive x-direction as our reference. Since the object is moving with a positive velocity (+3 m/s) after the impulse, we can conclude that the direction of the impulse is also in the positive x-direction.

Hence, the magnitude of the impulse is 0.6 kg·m/s in the positive x-direction.