Download MOMENTUM and Impulse

Momentum and
Impulse
Momentum (p)
A vector quantity equal to the product of
the mass of an object and the velocity of
an object.
The direction is always the same as the
direction of the velocity vector.
p = mv
m = mass (kg)
v = velocity (m/s)
p = momentum (kg m/s)
Ex: A 1.00 x 103 kg car is moving at
15 m/s [E]. What is its momentum?
Given:
m = 1.00x 103 kg
v = 15 m/s [E]
Find: p = ?
p = mv
= 1.00 x 103 kg (15 m/s [E])
=1.5 x 104 kg m/s [E]
Impulse
For a constant external force, it is the
product of the force acting on an object
and the time during which the force acts
on the object.
Ft =  p = m vf – m vi
Units = Ns = kg m/s

This equation is valid only if the force is
uniform over the time interval.
Impulse
Large forces can change momentum in a
very short time interval (hitting a baseball
or golf ball).
Small forces can cause the same change
in momentum as larger forces, but over a
longer time period (air bags in cars).
When an object bounces, a greater
impulse is required than when an object is
brought to a stop.
Ex. 1: A 0.15 kg baseball moving at 20. m/s is hit
by a bat, and leaves with the same speed, but
opposite direction. What is the change in
momentum of the ball?
Given: m = 0.15 kg
vi = 20. m/s
vf = -20. m/s
Find: Δp = ?
Δp = m vf – m vi
= 0.15 kg (-20. m/s-20. m/s)
= -6.0 kgm/s
Ex. 2: A 0.15 kg baseball moving at 20. m/s
is caught by the catcher. What is the change
in momentum of the ball?
Given: m = 0.15 kg
vi = 20. m/s
vf = 0
Find: Δp = ?
Δp = m vf – m vi
= 0.15 kg (0-20. m/s)
= -3.0 kgm/s
Ex: If the change in momentum in each case
occurs in 0.10 s, what force acted on the ball in
each case?
1.
2.
Given: Δp = -6.0 kgm/s
Given: Δp = -3.0 kgm/s
Δt = 0.10 s
Find: F = ?
Ft =  p
F =  p/ t
= (-6.0 kgm/s)/0.10s
= -60. N
Δt = 0.10 s
Find: F = ?
Ft =  p
F =  p/ t
= (-3.0 kgm/s)/0.10s
= -30. N