Example 6-28. A particle in a force field F=F(x, y, z )having a position
vector r =xˆe 1 +yˆe 2 +zˆe 3 moves according to Newton’s second law such that
F =ma =mdv
dt
or F dt =m dv.
An integration over the time interval t 1 to t 2 produces
∫t 2
t 1
F dt =m v(t 2 )−m v (t 1 ).
The quantity
∫t 2
t 1 F dt is called the linear impulse on the particle over the time interval
(t 1 , t 2 ).The quantity mv is called the linear momentum of the particle. The above
equation tells us that the linear impulse equals the change in linear momentum.
Example 6-29. In 10 seconds a particle with a mass of 1 gram changes velocity
from
v 1 = 6 ˆe 1 + 2 ˆe 2 + 7 ˆe 3 cm /s to v 2 =− 2 ˆe 1 +ˆe 3 cm /s.
What average force produces this change?
Solution: The average force over a time interval (t 1 , t 2 )is given by
Favg =^1
t 2 −t 1
∫t 2
t 1
F dt.
But the integral
∫t 2
t 1 F dt is the linear impulse and equals the change in linear mo-
mentum given by mv 2 −mv 1 .The average force is therefore
Favg =^1
10 [(−^2 ˆe^1 +ˆe^3 )−(6 ˆe^1 + 2 ˆe^2 + 7 ˆe^3 )]
=^1
5
[− 4 ˆe 1 −ˆe 2 − 3 ˆe 3 ] dynes.
Line Integrals of Scalar and Vector Functions.
An important type of vector integration is integration by line integrals. Let Cbe a
curve defined by a position vector
r =xˆe 1 +yeˆ 2 +zˆe 3 ,
