Calculus-Based Physics by Jeffrey W. Schnick - HTML preview

Chapter 26 Impulse and Momentum

As always, in applying Newton’s second law of motion, we start by drawing a free body

diagram:

m

F

a

In order to keep track of the vector nature of the quantities involved we apply Newton’s 2nd Law

in vector form (equation 14-1):

= 1

a

∑F

m

In the case at hand the sum of the forces is just the one force F , so:

1

a =

F

m

Solving for F, we arrive at:

F = a

m

multiplying both sides by ∆t we obtain

F∆t = ma t

∆

Given that the force is constant, the resulting acceleration is constant. In the case of a constant

acceleration, the acceleration can be written as the ratio of the change in v

that occurs during the

time interval ∆t, to the time interval ∆t itself.

v

= ∆

a

t

∆

Substituting this into the preceding expression yields:

v

∆

∆

F t = m

t

∆

∆t

F∆t = m v

∆

The change in velocity can be expressed as the final velocity v ′ (the velocity at the end of the

time interval during which the force acts) minus the initial velocityv (the velocity at the start of

the time interval): v

∆ = v ′−v . Substituting this into F∆t = m v

∆ yields

F∆t = (

m v ′− v )

which can be written as

F∆t = mv ′− mv

1

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