PHYSICS – Chp. 12: Momentum
Concept Review #3, Problems #3-7, 9, 10 (Worth 20 pts./ 18 pts available)
Concept Review
a.
The total momentum of a system is conserved (without outside forces). Individual objects can
increase or decrease momentum based on the velocity of the object.
b.
An object will have momentum if it has mass AND velocity.
c.
Correct.
d.
The units for momentum is kg∙(m/s)
e.
An object moving will have momentum.
f.
The momentum of two IDENTICAL objects moving at the same speed AND in the same direction
will have the same momentum.
g.
See letter f.
h.
Correct.
i.
If the momentum is positive and the acceleration is positive, then the momentum of the object
will increase.
j.
Momentum depends on mass AND velocity, so the faster object may not have a higher
momentum.
k.
Correct for objects with mass (if light is considered; its momentum depends on its energy).
Problems
3.
a.
p = m∙v
p = (0.04 kg)(50 m/s) = 2 kg∙m/s
b.
2 kg∙m/s = (0.15 kg)(v)
v = 13.33 m/s
1 mi = 1609.34 m
v = 29.82 mph
3600 s = 1 hr
4.
a.
The collision is inelastic. The cars stick together, and the momentum of each object is
not conserved (the objects’ momentums change).
b.
m1v1 + m2v2 = (m1 + m2)vf
(1200 kg ∙ 12 m/s) + (1000 kg ∙ -16 m/s) = (2200 kg)(vf)
vf = - 0.73 m/s
5.
m1v1 + m2v2 = (m1 + m2)(vf)
(75 kg)(-3 m/s) + (70 kg)(2 m/s) = (75 kg + 70 kg)(vf)
-85 kg∙m/s = (145 kg)vf
-0.586 m/s = vf
Players are moving 0.586 meters per second towards the East.
6.
Consider law of conservation of momentum
ptotal initial = ptotal final
m1v1 + m2v2 = m1v3 + m2v4
(4.5 kg)(5 m/s) + (5 kg)(-7 m/s) = (4.5)(-8.5 m/s) + (5 kg)(v4)
-12.5 kg∙m/s = (-38.25 kg∙m/s) + (5 kg)(v4)
25.75 kg∙m/s = (5 kg)(v4)
v4 = 5.15 m/s
7.
a.
I = Favg ∙ ∆t
I = (20 N)(2 s) = 40 N∙s
b.
numerically speaking, the impulse is the same as change in momentum, so
∆p = 40 kg∙m/s
9.
a.
Moment of Inertia of sphere = (2/5)m∙r2
I = (2/5)(0.5 kg)(0.1 m)2 = 0.002 kg∙m2
*found on pg. 255
b.
L = I∙ω
*from Chp. 8, v = ω∙r , so
3 m/s = ω (0.1 m)
ω = 30 rad/s
L = (0.002 kg∙m2)(30 rad/s) = 0.06 (kg∙m2)/s
c.
p = mv
p = (0.5 kg)(3 m/s) = 1.5 kg∙m/s
10.
a.
L = I∙ω = (1.5 kg∙m2)(10 rad/s) = 15 (kg∙m2)/s
b.
Considering the Law of Conservation of Angular Momentum,
L – L0 = 0
-or-
L = L0
15 (kg∙m2)/s = (3 kg∙m2)(ω)
ω = 5 rad/s