Transcript Momentum: The quantity of motion of a moving body

Momentum:
The quantity of motion of a moving
body, measured as a product of its
mass and velocity
Momentum
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Represented by the variable P.
P = mv, where m = mass, kg & v = velocity, m/s
Momentum is conserved; it cannot be created or destroyed.
Any changes in momentum of a body will generate changes in
momentum elsewhere in its system.
Momentum is a different quantity from kinetic energy in that
its velocity is not squared.
If two objects collide, their respective momentums will both
change.
A collision is a classic type of demonstration of momentum.
Two types of collisions: Elastic and non-elastic
Elastic: Objects bounce off each other with no energy loss.
Non-elastic: Objects stick together when they collide.
• Changes in momentum are associated with applied forces. Remember Newton’s 1st
Law: “An object at rest stays at rest and an object in motion stays in motion with
the same speed and in the same direction unless acted upon by an unbalanced
force.”
• When two objects exchange momentum (collide), we represent it mathematically
as:
Ptotal = (P1i + P2i) = (P1f + P2f); (m1v1i)+ (m2v2i)= (m1v1i)+ (m2v2i )
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Where: P1 = momentum of 1st object; P2 = momentum of 2nd object,
i = initial, before collision; f = final, after collision
Note: velocities change but masses do not.
When an outside force is applied to an object in motion, its momentum is changed.
(Force)(time) = (mass)(change in velocity): F t = m Δv
The quantity Force • time is known as impulse. Impulse = Change in momentum
And since the quantity m • v is the momentum, the quantity m•Δv will be
the change in momentum.
Recall that velocity is a VECTOR, so it has both magnitude and direction.
Example of Impulse: When a batter hits a pitch, an impulse is applied to the bat
and the ball, which changes their respective momentums.
Example Problems
1) What is the magnitude of momentum of a 1.5 kg mass traveling at a rate of
5.0 m/s?
P = mv = (1.5 kg)(5.0 m/s) = 7.5 kg m/s
2) A ball with mass 50 g is thrown with a velocity of 7.0 m/s at a stationary, 10 g
aluminum can. The ball strikes the can and continues forward at 4.7 m/s
(elastic collision). What is the velocity of the can after the ball strikes it?
Pib = (0.050 kg)(7.0 m/s) = 0.35 kg m/s; Pic = 0 (the initial velocity = 0)
Pfb = (0.050 kg)(4.7 m/s) = 0.235 kg m/s; Pfc = (0.010 kg) vfc
0.35 + 0 = 0.235 + 0.010 vfc
(0.35 ‒ 0.235) / 0.01 = vfc
vfc = 11.5 m/s
3) Next, consider problem #2, except this time the ball and the can stick
together upon colliding (non-elastic collision). What is the velocity for the
combined objects after the collision?
Pib = (0.050 kg)(7.0 m/s) = 0.35 kg m/s; Pic = 0 (the initial velocity = 0)
Ptot = (0.05 + 0.01) kg vf-tot (Note: The two masses are now combined.)
0.35 kg m/s = 0.06 kg vf-tot
vf-tot = 5.8 m/s
4) A football with mass = 425 g (yep, I looked this up) is kicked from a tee such
that its horizontal velocity = 10 m/s.
a. What horizontal impulse was applied to the ball?
Impulse = m Δv = (0.425 kg)(10 – 0) m/s = 4.25 kg m/s
b. If this impulse’s time duration = 0.20 s, what force was applied?
m Δv = F t; t = (4.25 kg m/s) / 0.20 s = 21.25 kg m2/s2 = 21.25 N