Question:

A 0.5 kg object is initially at rest on a frictionless surface. Suddenly, a constant force of 10 N is applied to the object.

a) Calculate the impulse experienced by the object over a time interval of 2 seconds.

b) Calculate the final velocity of the object after the force is applied.

c) Calculate the momentum of the object after the force is applied.

Answer:

a) To calculate the impulse experienced by the object, we can use the equation:

Impulse (J) = Force (F) × time interval (Δt)

Given:

Force (F) = 10 N

Time interval (Δt) = 2 s

Substituting the given values into the equation:

Impulse (J) = 10 N × 2 s = 20 N·s

Therefore, the impulse experienced by the object over a time interval of 2 seconds is 20 N·s.

b) To calculate the final velocity of the object, we can use the equation:

Impulse (J) = Change in momentum (Δp)

The impulse is equal to the change in momentum of the object. Using the equation:

Δp = m × Δv

Where: Δp = change in momentum m = mass of the object Δv = change in velocity

Since the object is initially at rest, the initial velocity (u) is 0. Therefore, we have:

Δv = v - u = v - 0 = v

Rearranging the equation:

Δp = m × Δv 20 N·s = 0.5 kg × Δv

Solving for Δv:

Δv = 20 N· s / 0.5 kg = 40 m/s

Therefore, the final velocity of the object after the force is applied is 40 m/s.

c) Momentum (p) is defined as the product of an object's mass and its velocity. So, we can calculate the momentum using the equation:

p = m × v

Given:

Mass (m) = 0.5 kg Velocity (v) = 40 m/s

Substituting the given values into the equation:

p = 0.5 kg × 40 m/s = 20 kg·m/s

Therefore, the momentum of the object after the force is applied is 20 kg·m/s.