Name______________________________Block_____Date_______________________
Ch. 7 Momentum
Mrs. Peck
7.1 Momentum
momentum- the product of the mass and velocity of an object
-is a vector quantity: has magnitude and direction
-a measure of how difficult it is to stop a moving object
-is “moving mass” or “inertia in motion”
- mass in motion
-momentum depends on both mass and velocity
momentum = (mass) ( velocity)
symbol is p
SI unit is kg.m/ s
p=mv
p~m
p~v
if double mass then double momentum
if double velocity then double momentum
fig. 7.1
at rest vel. is 0m/ s ----> p = 0 kg.m/ s
change in momentum ∆p = m∆v if velocity is changing.... a net force acted on the object
if velocity is changing....object is accelerating (pos. or neg.)
∆p = m∆v
∆v=(vf - vo)
? 1. a What is the momentum of a 10.5 kg bowling ball rolling at 3 m/ s?
b. What would the momentum be if the mass of the bowling ball were doubled and its velocity
still was 3 m/ s?
c. What would be the momentum if the mass stayed at 10.5 kg and the velocity halved?
d. What would be the momentum if the mass and velocity were doubled?
2. What is the momentum of a 60 kg halfback moving forward at 8 m/ s ?
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3. What is the momentum of a 1500 kg car moving northward at 25 m/ s?
4. Tiger Woods hits a 0.05 kg golf ball, giving it a speed of 75m/ s. What is the change in
momentum of the ball?
5. Will a large truck always have more momentum than a roller skate?
7.2 Impulse Changes Momentum
impulse- the product of force and time interval during which a force acts
-the product of the force exerted on an object & the time interval during wh/ it acts
- apply force...changes velocity........obj accelerates........changes momentum
(positive or negative)
an object with momentum is hard to stop......must apply a force over period of time to stop it
the more the momentum ....the harder it is to stop....the more force or longer amt of time (or both) to stop it
as force is applied to object for a given amt of time....its velocity changes....its momentum changes
if force is in same dir. as object’s motion...it speed up (positive acceleration).....∆ momentum
if force is in opp. dir. as object’s motion....it slows down (negative acceleration).....∆ momentum
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7.2 Impulse Changes Momentum
-apply brief force: ∆’s momentum....apply force for extended time: greater ∆ in momentum
impulse- the product of force and time interval during which a force acts
impulse = (force)(elapsed time)
symbol is J
SI unit is N. s
J=F∆t
impulse = change in momentum
J=∆p
impulse ~ ∆ momentum (impulse does NOT equal momentum!)
increase impulse exerted then increase change in momentum
J=∆p
F∆t = m∆v
(N)(s) =kg. m/s
.
(kg m/s2)(s)= kg. m/s
F∆t=m∆v
a= F
m
the impulse given to an object is equal to the ∆ in momentum
“unit analysis”
a = ∆v
t
a=a
F = ∆v
m
t
cross-multiply
Ft = m∆v
* the same change in momentum may be result of 1. large force exerted for short time
2. small force exerted for long time
impulse is the thing that you do
momentum is the thing that you see
momentum & impulse
impact- is a force : unit N
time during wh/ your momentum is brought to 0 kg m/s
impulse- is a impact force x time : unit N. s
fig. 7.3 if increase time of impact.....decrease force of impact....decrease acceleration
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fig. 7.4 if decrease time of impact....increase force of impact...
fig. 7.5 if increase time of impact...decrease force of impact
if decrease time of impact...increase force of impact
if drop something and it lands on different surfaces.: cement and grass
momentum is same same mass & velocity
p = m∆v
if momentum is same then the impulse for both is same
if impulse is the same then only the time of impact and force of impact are ∆g
if increase time of impact then decrease force of impact
if decrease time of impact then decrease force of impact
impulse will not change if increase time of impact AND decrease force of impact
? When a dish falls, will the impulse be less if it lands on a carpet than if it lands on a hard floor?
Why
? If a boxer is able to make the impact time five times longer by “riding” with the punch, how much
will the force of impact be reduced? Why?
7.3 Bouncing
Bouncing: impulses are greater when obj. bounces
magnitude of impact force depends on impact time
impulse to bring obj. to a stop
+ impulse to “throw” back again
7.4 Conservation of Momentum
to accelerate an object......must exert net force upon it
to change the momentum of an object..... must exert an impulse upon it
*the impulse or force must be exerted on obj. by something outside of object or system
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fig. 7.8 -net momentum before and after firing is 0 kg. m/s
-momentum of cannonball and momentum of canon ...= & opp....cancel out
-no net outside force acted on cannonball-cannon system...no net impulse on
system....no net change in momentum
momentum is a vector quantity ........has both magnitude and direction
momentum can be canceled by equal and opposite momentum
*if no net force or net impulse acts on a system, the momentum of that system cannot change
-momentum of a system cannot change unless it is acted on by external forces
-a system will have the same momentum before some internal interaction as it has after the interaction occurs.
conserved- term applied to a physical quantity, such as momentum, energy, or electric
charge, that remains unchanged during interactions
law of conservation of momentum- in the absence of a net external force, the momentum of
an object or system of objects is unchanged
total momentum in a system remains the same if no external forces act on the system
Object 1 and object 2 collide: the total momentum of the two objects before the
collision is equal to the total momentum of the two objects after the collision. That is,
the momentum lost by object 1 is equal to the momentum gained by object 2.
The total momentum of a collection of objects (a system) is conserved....that is,the
total amount of momentum is a constant value (total momentum does not change)
collision of object 1 & object 2:
forces are = & opp
time of impact is the same
impulses are the = & opp direction
impulses = change of momentum
impulse are the same, then changes of momentum are the same
changes of momentum are the equal & opp direction
F1 = F2
t1 = t 2
F1 t1 = - F2 t2
J = ∆p
∆p1 = ∆p2
m1 ∆v1 = - m2 ∆v2
momentum lost by object 1 is gained by object 2
one object slows down and loses momentum & the other object speeds up and gains momentum
the total momentum of the two objects is the same before and after the collision
the total momentum of the system (collection of two objects) is conserved
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elastic collision - the property of a solid wherein a change in shape is experienced
when a deforming force acts on it, with a return to its original shape
when the deforming force is removed.
elastic collision: a collision in which objects collide and bounce apart with no energy loss
change or momentum is always conserved
NOTE: if object 1 and object 2 are moving in opposite directions.....then must assign
one object with a positive velocity and the other object with a negative
velocity
Use the following subscripts:
o - original
f - final
1 - for object 1
2 - for object 2
total momentum before collision = total momentum after collision
(original)
(p1o
+
(final)
p2o) = ( p1f
+
p2f )
m 1ov1o + m 2ov2o = m 1f v1f + m 2fv2f
Isolating Variables: if given all but one variable, must rearrange formula to solve for the variable
solve for v1f : must move all variables to one side of equation except v1f
(PB #7 p58)
step 1: move (m2fv2f) to the left by subtracting it from both sides
m1ov1o + m 2ov2o - (m2fv2f) = m 1f v1f + m 2fv2f - (m2fv2f)
m1 ov1 o + m2 ov2 o - (m2fv2f) = m1f v1f
step 2: get rid of m1f from m1f v1f by dividing both sides by m1f leaving v1f only
(m1 ov1 o + m2 ov2 o -m2fv2f) = m1f v1f
m1f
m 1f
(m1 ov1 o + m2 ov2 o - m2fv2f) = v1f
m1f
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inelastic collision - term applied to a material that does not return to its original
shape after it has been stretched or compressed (also called plastic)
a collision in which objects collide & some mechanical E transformed into heat
3 situations of inelastic collisions:
1. 2 objects collide and get stuck together
2. 2 objects stuck together and separate
3. 2 objects collide and one or both are deformed (E lost to heat)
momentum is conserved:
2 objects collide and get stuck together
obj 1
+
obj 2 = obj 1 obj 2
total momentum before collision = total momentum after collision
(original)
(p1o
+
(final)
p2o) =
(pf)
m 1ov1o + m 2ov2o = (m1 + m 2 ) vf
Isolating Variables: given all variables except for one
How to solve for vf: move all variables to one side of equation except vf
(PB #6 p57, #10 p 59)
step 1: move (m1 + m2 ) to the left by dividing it from both sides to solve for vf
m 1ov1o + m 2ov2o = (m1 + m 2 ) vf
(m1 + m 2 )
(m1 + m 2 )
m 1ov1o + m 2ov2o = vf
(m1 + m 2 )
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Inelastic collisions:
2 objects stuck together and then separate
obj. 1
obj. 2
=
+
obj. 1
obj. 2
total momentum = total momentum
before collision
after collision
(original)
(final)
(m1 + m 2 ) vo = m 1fv1f + m 2fv2f
Isolating Variables: given all variables except for one.
solve for v1f: must move all variables except v1f to one side of equation
(PB #7 p57)
step 1: move (m2fv2f) to the left by subtracting it from both sides
(m1 + m 2 ) vo - (m2fv2f) = m 1fv1f + m 2fv2f - (m2fv2f)
(m1 + m 2 ) vo - (m2fv2f) = m 1fv1f
step 2: must move m1f from m1fv 1f by dividing both sides of equation by m1f
(m1 + m 2 ) vo - (m2fv2f) = m 1fv1f
m 1f
m 1f
(m1 + m 2 ) vo - (m2fv2f) = v1f
m 1f