Transcript Linear Impulse − Momentum

KINE 3301
Biomechanics of Human Movement
Linear Impulse − Momentum
Chapter 8
Definitions
• Momentum: mass x velocity (units kg∙m/s)
• Conservation of Linear Momentum – The total linear
momentum of a system of objects is constant if the net
force acting on a system is zero.
• Elastic Collision: The objects collide and rebound.
• Inelastic Collision: The objects collide and stick together.
• Impulse (units N∙s)
– Constant force: Average force x time.
– Non-Constant force: Area under the force – time curve.
• Impulse – Momentum: The impulse is equal to the change
in momentum.
Equations
p=mv
Linear Momentum
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = 𝑚𝐴 𝑣𝐴 + 𝑚𝐵 𝑣𝐵
Elastic Collisions
Inelastic Collisions
𝑚𝐵 𝑢𝐵
𝑣𝐴 =
𝑚𝐴
𝑚𝐴 𝑢𝐴
𝑣𝐵 =
𝑚𝐵
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = (𝑚𝐴 +𝑚𝐵 )𝑣
𝑡1
Impulse
𝐽 = 𝐹∆𝑡
𝐽=
𝐹 𝑑𝑡
𝑡0
𝑡1
Impulse−Momentum
𝐹 ∆𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝐹 𝑑𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
𝑡0
Linear Momentum
• The linear momentum (p) of an object is the product
of it’s mass (m) and velocity (v). The units for linear
momentum are kg∙m/s.
m = 2 kg
p=mv
p = (2 kg) (3 m/s)
p = +6 kg∙m/s
v = 3 m/s
p
The vector for linear momentum points
in the same direction as the velocity.
Conservation of Linear Momentum
• The total linear momentum of a system of objects is
constant if the net force acting on a system is zero.
• The total linear momentum is defined by:
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = 𝑚𝐴 𝑣𝐴 + 𝑚𝐵 𝑣𝐵
𝑢 is the initial velocity (before collision)
𝑣 is the final velocity (after collision)
𝑚 is the mass of the object
Collision Classifications
• Collisions are classified according to whether the
kinetic energy changes during the collision.
• The two classifications are elastic and inelastic.
• In an elastic collision the total kinetic energy of
the system is the same before and after the
collision.
• In an a perfectly inelastic collision the total kinetic
energy is still conserved but the two objects stick
together and move with the same velocity.
Conservation of Linear Momentum
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = 𝑚𝐴 𝑣𝐴 + 𝑚𝐵 𝑣𝐵
The equation above is usually rearranged for
elastic and inelastic collisions as follows:
𝑚𝐴 𝑢𝐴
𝑣𝐵 =
𝑚𝐵
Elastic
Collisions
𝑚𝐵 𝑢𝐵
𝑣𝐴 =
𝑚𝐴
Inelastic
Collisions
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = (𝑚𝐴 +𝑚𝐵 )𝑣
Two billiard balls collide in a
perfectly elastic collision. Ball A
has a mass of 0.8 kg and an
initial velocity (uA) of 3 m/s, ball
B has a mass of 0.3 kg and an
initial velocity (uB) of −2 m/s,
determine the velocity of each
ball after the collision.
𝑚 𝐵 𝑢𝐵
𝑣𝐴 =
𝑚𝐴
(.3𝑘𝑔)(−2𝑚/𝑠)
𝑣𝐴 =
.8𝑘𝑔
𝑣𝐴 = −0.75𝑚/𝑠
𝑚𝐴 𝑢𝐴
𝑣𝐵 =
𝑚𝐵
(.8𝑘𝑔)(3𝑚/𝑠)
𝑣𝐵 =
.3𝑘𝑔
𝑣𝐵 = 8.0 𝑚/𝑠
Two clay objects collide in
an inelastic collision, object
A has a mass of 0.8 kg and
an initial velocity (uA) of 4
m/s, object B has a mass of
0.4 kg and an initial velocity
(uB) of −2 m/s, determine
the final velocity of A and B.
𝑚𝐴 𝑢𝐴 + 𝑚𝐵 𝑢𝐵 = (𝑚𝐴 +𝑚𝐵 )𝑣
.8 𝑘𝑔
𝑚
4
+ .4 𝑘𝑔
𝑠
𝑚
−2
= (.8 𝑘𝑔 + .4 𝑘𝑔)𝑣
𝑠
𝑣 = 2 m/s
Computing Impulse
Constant Force
𝐽 = 𝐹∆𝑡
Impulse = Average Force x time
Non−Constant Force
𝑡1
𝐽=
𝐹 𝑑𝑡
𝑡0
Impulse = area under force-time curve
Impulse
Impulse (J) is defined as product of an average force (𝐹) and
time (∆𝑡), or the area underneath the force time graph. The
units for impulse are N∙s.
𝐽 = 𝐹∆𝑡
𝑡1
𝐽=
𝐹 𝑑𝑡
𝑡0
Computing Impulse using Average Force
Compute the Impulse (J) for the force shown below with an
average force (𝐹) 95.6 N and time (∆𝑡) of 0.217 s.
𝐽 = 𝐹∆𝑡
𝐽 = 95.6𝑁 .217𝑠
𝐽 = 20.76 𝑁 ∙ 𝑠
Impulse-Momentum
The impulse – momentum relationship is derived from
Newton’s law of acceleration.
𝐹 = 𝑚𝑎
𝑉𝑓 − 𝑉𝑖
𝑎=
Δ𝑡
𝑉𝑓 − 𝑉𝑖
𝐹=𝑚
Δ𝑡
𝐹 ∆𝑡 = 𝑚𝑉𝑓 − 𝑚𝑉𝑖
Impulse = change in momentum
A soccer player imparts the force shown below on a soccer ball with a mass of
0.43 kg and an initial velocity (Vi) of 0.0 m/s. After the force was applied the ball
had a final velocity (Vf) of 23.02 m/s. The average force F of 90.8 N was applied
for 0.109 s. Compute the impulse using both average force and the change in
momentum.
A softball player imparts the force shown below on a softball with a mass
of 0.198 kg and an initial velocity (Vi) of 0.0 m/s. After the force was
applied the ball had a final velocity (Vf) of 50.51 m/s. The average force F
of 31.74 N was applied for 0.315 s. Compute the impulse using both
average force and the change in momentum.
60
60
Force (N)
Velocity (m/s)
50
40
40
30
30
20
20
10
10
0
0
0
0.05
0.1
0.15
0.2
Time (s)
0.25
0.3
0.35
Velocity (m/s)
Force (N)
50