Question

Repeat the preceding problem, but including a drag force due to air of ( vec{f}_{{drag}} = -bvec{v} ).

Reference Problem:
A ball of mass m is dropped. What is the formula for the impulse exerted on the ball from the instant it is dropped to an arbitrary time τ later? Ignore air resistance.

a) ( J = -mgtau - bvec{v} )

b) ( J = -mtau - bvec{v} )

c) ( J = -mg - bvec{v}tau )

d) ( J = -mvec{v} - btau )

Answer 1

The correct formula for the impulse exerted on the ball with gravity and air resistance accounted for is option (c): (J = -mgτ - bττ), summing the gravitational and the velocity-proportional drag effects over time.

Step-by-step explanation:

The student is asking a physics question about the impulse on a ball including a drag force due to air. The correct formula for the impulse exerted on the ball from the moment it is dropped to an arbitrary time τ later, accounting for both gravity and air resistance, must consider both the force of gravity (which is a constant –mg) and the drag force (which is proportional to velocity and therefore changes over time).

Therefore, the correct option would imply a sum of two terms: one representing the constant gravitational force over time (–mgτ) and the other representing the cumulative effect of the drag force (which is dependent on velocity).

Given the options presented and considering the terms correctly, the correct option is: (c) (J = -mgτ - bττ). The formula captures the constant force due to gravity (-mgτ) and the effect of the drag force over time (-bττ), with the drag force being proportional to the velocity of the ball, which is in the direction opposite to the motion (hence the negative sign).