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Ch
9 Momentum
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Updated 5/10/04
NEWTON'S
3rd LAW: FOR EVERY ACTION THERE IS
AN EQUAL AND OPPOSITE REACTION
THAT IS A SINGLE FORCE CAN NEVER EXIST. WHENEVER
TWO OBJECTS INTERACT EXACTLY EQUAL AND OPPOSITE FORCES APPEAR.
THUS THE IMPULSE, Ft, GIVEN TO ONE
OBJECT MUST BE EXACTLY THE SAME AS
OBJECT A OBJECT B
THE IMPULSE GIVEN TO THE SECOND
Ft
= -Ft
BUT IN OPPOSITE DIRECTION.
MΔV =
-MΔV
Algebraic
Analysis of Collisions
THERE
ARE 3 TYPES OF COLLISIONS.
ELASTIC:
KE BEFORE = KE AFTER (RARE), NO DEFORMATION
INELASTIC:
KE BEFORE > KE AFTER, OBJECTS ARE DEFORMED.
PERFECTLY INELASTIC KE BEFORE > KE AFTER, V1' = V2'
WHEN TWO THINGS COLLIDE, THEIR MOMENTUM BEFORE AND AFTER THE COLLISION IS THE
SAME.
M1V1 + M2V2
= M1V1' + M2V2'
SOLVING FOR V2': V2' = V2 +
M1(V1-V1')/M2
CASE 1, V1 = V2 = 0, ROCKETS & GUNS/CANNONS:
V2' = V2 + M1(V1-V1')/M2--->V2' = -M1V1'/M2
M1V1'2 + M2V2'2
M1V1'2 + M2V2'2
%K.E. REMAINING = --------------- = --------------- = INFINATE---> Not a
collision
M1V12 + M2V22
0
but a launch
***************************************************************************
CASE 2, V2 = 0, COLLISION WITH SATIONARY OBJECT: V2' = M1(V1-V1')/M2
M1V1'2 + M2V2'2
M1V1'2 + M2V2'2
M1V1'2 M1(V1-V1')
2
%K.E. REMAINING = --------------- = --------------- = ------ + ------------
M1V12 + M2V22
M1V12
M1V12
M2V12
THIS IS ELASTIC ONLY IF M1 = M2, AND V1' = 0, OBJECTS TRADE SPEEDS.
***************************************************************************
CASE 3, PERFECTLY INELASTIC, WITH SATIONARY OBJECT: V2 = 0, V1'= V2' = V'
= ?
SOLVING M1V1 + M2V2 =
M1V1' + M2V2' FOR V'--->V' = M1V1/(M1 + M2)
M1V1'2 + M2V2'2
(M1 + M2)V'2
M1
%K.E. REMAINING = --------------- = ------------ = -------
M1V12 + M2V22
M1V12
M1 + M2
WHICH IS INDEPENDENT OF V1.
IF M1 = M2, %K.E.rem = 50%
IF M1 is much larger than M2 (M1 >> M2, Ex: M1 =100, M2 =1), %K.E.rem = 100%
IF M1 is much smaller than M2 (M1<< M2 Ex: M1 =1, M2 =100), %K.E.rem = 0%
FOR
THE MOMENTUM OF A SYSTEM TO CHANGE AN EXTERNAL FORCE MUST ACT ON IT.
AN INTERNAL FORCE CANNOT CHANGE THE TOTAL MOMENTUM OF A SYSTEM.
1
mi/hr = 0.447 m/s = 1.467 ft/s
1 lb = 4.445 N 1
kg = 2.2 lb
Note: Lower case m or v indicates a small mass or velocity
Upper case M or V indicates a large mass or velocity.
1
a] VERBALLY & ALGEBRAICALLY DEFINE IMPULSE & MOMENTUM.
Momentum = Mass x Velocity (a vector)
p
= mV
Impulse = Applied Force x Contact Time
= Ft
b] WHICH HAS MORE MOMENTUM A TRUCK AT REST OR A ROLLING
SKATEBOARD?
Truck: p = Large Mass x Zero Velocity
= 0 momentum
Skateboard: p = Small mass x Small
Velocity = small momentum <---Ans
c] UNDER WHAT CONDITIONS CAN A BICYCLE & A CAR HAVE THE
SAME NON-ZERO MOMENTUM?
A fast bicycle can have the same momentum
as a slow car.
Bicycle = Car
Numerically Bicycle = Car
mV = Mv
10 kg x 100 m/s = 1000 kg x 1 m/s
d] USE NEWTON'S 2nd LAW TO DERIVE THE IMPULSE-MOMENTUM
EQUATION.
2nd Law:
F = ma
Substituting ΔV/t
for a:
F = mΔV/t
Multiply both sides by t:
Ft = mΔV
UNITS:
N sec = kg m/s
e] STATE THE EQUATION VERBALLY 2 WAYS.
Outside Force(F) x Contact time(t) causes(=) a momentum change(mΔV)
Impulse(Ft) causes(=) a momentum change(mΔV)
Note: IMPACT
is the force on an object: F
IMPULSE is the force x contact time:
Ft
2
a] Compare Collisions with Launches.
b] HOW DO AIR BAGS AND SEAT BELTS WORK?
c] WHY SHOULD A BOXER RIDE WITH A PUNCH?
d] WHY SHOULD A BATTER FOLLOW THROUGH WITH THE SWING? (CAUSE
& EFFECT)
Collisions
start with a given needed momentum change.
In a car crash the needed momentum change is determined by your mass and speed.
That is mΔV
= constant.
Applying this to Ft = m (Vf - Vi):
Ft = constant ---> F = constant/t
Or, the applied force is inversely proportional to the contact time.
If you increase the contact time & you decrease the force.
Ex: Seat
belts and air bags increase the contact time to decrease the force.
Boxers ride with the punch to decrease the force by increasing the
contact time.
Ft
= mΔV
Collisions
12x1 = 12
Start with a given initial & final momentum mΔV (12)
6x2
= 12
Increase the contact time to decrease the force.
4x3
= 12
Ex:
Seat belts
2x6
= 12
Launches
start with a given force, F = constant.
Applying this to Ft = m (Vf - Vi)--->
Momentum Change = Constant x time
If you increase contact time then you increase momentum change.
Ex: Follow
through when batting or throwing a ball increases contact time and the balls
momentum.
Ft
= mΔV
Launches
12x1 = 12
Start with a given force (12).
12x2 = 24
Increase the contact time to increase the momentum
12x3 = 36
Ex:
Throwing a ball, follow through when batting
3
a] ANALYZE CATCHING & THROWING A BALL WHILE RIDING A SKATEBOARD.
You move back once when you catch the ball and again when you throw it.
Thus, your momentum change is doubled.
This simulates bouncing.
Math:
Vf = -Vi--->mΔV = m(Vf – Vi) = m(-Vi – Vi) = -2mVi
b] THE EFFECT OF BOUNCING is to double THE CHANGE IN
MOMENTUM
c] A GAS TURBINE bounces the gas thus doubling the wheel’s
momentum change.
4 540 N IS USED TO STOP AN 65 kg OBJECT MOVING AT 175 m/s.
a] DRAW & LABEL THE DIAGRAM.
+----+
540 N---->|Boat|<----175 m/s
+----+
b] HOW LONG WILL IT TAKE TO BRING THE OBJECT TO A FULL STOP?
F
= + 540 N
Equation: Ft =
m(Vf - Vi)
m = 65 kg
Solve for t: t = m(Vf -
Vi)/F
Vi = - 175 m/s
Sub numbers: t = 65 kg(0 -
(-175m/s))/540 kg m/s/s)
Vf = 0 m/s
t
= 21.1 s
t = ?
5
121.5 lb IS USED TO STOP AN 143 lb PERSON MOVING AT 78.2 mi/h.
a] DRAW AND LABEL THE DIAGRAM.
+------+
121.5 lb---->|Person|<----78.2
mi/h
|143 lb|
+------+
b] HOW LONG WILL IT TAKE TO BRING THE OBJECT TO A FULL STOP?
Convert from English to Metric Units
F = 121.5 lb x 4.445N/lb = 540 N
m = 143 lb x 1 kg/2.2 lb = 65 kg
V = 78.2 mi/h x 0.447 m/s/mi/h = 35 m/s
+------+
540 N---->|Person|<----35.0
m/s
| 65 kg|
+------+
F
= + 540 N
Equation: Ft =
m(Vf - Vi)
m = 65 kg
Solve for t: t = m(Vf -
Vi)/F
Vi = - 35 m/s
Sub numbers: t = 65 kg(0 -
(-35 m/s))/540 kg m/s/s)
Vf = 0 m/s
t
= 4.21 s
t = ?
c] HOW MUCH FORCE (lb) WILL IT TAKE TO STOP IT IN 15 s?
F
= ? lb
Equation: Ft =
m(Vf - Vi)
m = 65 kg
Solve for F: F
= m(Vf - Vi)/t
Vi = - 35 m/s
Sub numbers: F
= 65 kg(0 - (-35m/s))/15 s)
Vf = 0 m/s
F = 151.7 N
t = 15
s
Convert: F = 151.7 N x 1 lb/4.445 N = 34.1 lb
d] HOW LONG WILL IT TAKE TO STOP IF A 250 lb FORCE IS USED?
Convert from English to Metric Units
F = 250 lb x 4.445 N/lb = 1,111 N
Equation: Ft =
m(Vf - Vi)
m = 65 kg
Solve for t: t = m(Vf -
Vi)/F
Vi = - 35 m/s
Sub numbers: t = 65 kg(0 -
(-35 m/s))/1,111 kg m/s/s)
Vf = 0 m/s
t
= 2.05 s
t = ?
6
A CAR CARRYING A 100 lb CHILD IS MOVING AT 30 mi/h GETS INTO AN ACCIDENT.
IT STOPS IN 1 CAR LENGTH (20 ft).
FIND:
a] THE PASSENGER'S MASS.
100 lb x 1 kg/2.2 lb = 45.45 kg
b] THE CAR'S SPEED--m/s & ft/s.
30 mi/h * 0.445 m/s/mi/h = 13.35 m/s
30 mi/h * 1.467 ft/s/mi/h = 44.0 ft/s
c] THE TIME TO STOP.
D = (Vf + Vi)t
2
20 ft = (0
+ 44.0 ft/s)t ----> 20 = 22t ---> t = 20/22 = 0.909 s
2
d] THE FORCE EXERTED ON THE PASSENGER-lbs.
F
= ? lb
Equation: Ft =
m(Vf - Vi)
m = 45.45 kg
Solve for F: F = m(Vf -
Vi)/t
Vi = -13.35 m/s
Sub numbers: F =
45.45 kg(0 - (-13.35m/s))/0.909 s)
Vf = 0 m/s
F
= 667.5 N
t = 0.909
s
Convert: F = 667.5 N x 1 lb/4.445 N = 150.2 lb
CONSERVATION
OF MOMENTUM
7
STATE NEWTON'S 3rd LAW IN WORDS & AS AN EQUATION.
NEWTON'S 3rd LAW:
FOR EVERY ACTION THERE IS AN EQUAL AND OPPOSITE REACTION
Force on Object A = - Force on Object B
Fa = - Fb
THAT IS A SINGLE FORCE CAN NEVER EXIST.
WHENEVER TWO OBJECTS INTERACT EXACTLY EQUAL AND OPPOSITE FORCES
APPEAR.
THUS THE IMPULSE, Ft, GIVEN TO ONE
OBJECT MUST BE EXACTLY THE SAME AS
OBJECT A
OBJECT B
THE IMPULSE GIVEN TO THE SECOND
FaT
=
-FbT
BUT IN OPPOSITE DIRECTION.
MaΔVa
=
-MbΔVb
Note the contact times are equal, Ta = Tb = T
8
a] YOU CANNOT TOUCH WITHOUT being touched
b] FORCES ALWAYS ACT in pairs
c] ACTION & REACTION ALWAYS ACT ON DIFFERENT
OBJECTS.
d] WHEN WALKING YOUR FOOT PUSHES AGAINST THE
FLOOR, & THE FLOOR pushes
against you!
e] HOW CAN A ROCKET BE PROPELLED ABOVE THE
ATMOSPHERE WHERE THERE IS NO AIR TO
"PUSH AGAINST"?
The hot gases from the rocket push against the rocket
f] CAN YOU HIT A FEATHER IN MIDAIR WITH A FORCE OF 10 lbs?
WHY?
You cannot hit a feather
with a 10 lb force because the feather
cannot hit back with a 10 lb force.
9
a] DERIVE THE CONSERVATION OF MOMENTUM LAW FROM NEWTON'S 3rd LAW.
3rd Law:
Fa = -Fb
2nd Law: F =
ma--->
MaAa = - MbAb
Let V’ = Vf & let V = Vi
a = (V'-V)/t---> Aa = (Va’ - Va)/t
Substituting MaAa = - MbAb---> Ma(Va’ - Va)/t = -Mb(Vb’
- Vb)/t
Multiply by t:
Ma(Va’ - Va) = -Mb(Vb’ - Vb)
Algebra---->
MaVa + MbVb = MaVa’ + MbVb’
b] WRITE IT VERBALLY.
THE TOTAL(+) MOMENTUM BEFORE A COLLISION = TOTAL(+) MOMENTUM
AFTER A COLLISION
c] STATE THE ASSUMPTIONS.
No external forces, no friction, head on
collisions.
d] WRITE THE EQUATION FOR THE THREE BODY PROBLEM.
MaVa + MbVb +McVc = MaVa’ + MbVb’ +
McVc’
e] DEFINE & GIVE EXAMPLES OF ELASTIC & INELASTIC
COLLISIONS.
ELASTIC:
Both Momentum & Kinetic Energy
are Conserved, (RARE),
Characterized by bouncing & no
deformation (bending or breaking) occurs.
Ex: pool ball collisions
INELASTIC: Only
momentum is conserved. Some Kinetic Energy is converted
into heat, sound, and deformation (KE BEFORE
> KE AFTER).
Characterized by sticking.
Ex: clay ball collisions.
Most collisions are part elastic and
part inelastic.
See algebraic analysis of collisions above.
f] DRAW & LABEL DIAGRAM FOR A TOSSED BOMB THAT BREAKS IN
TWO.
FOR
THE FOLLOWING
a] DRAW BEFORE & AFTER DIAGRAMS.
b] PREDICT THE ANSWER
c] FIND THE UNKNOWN SPEED.
d] FIND THE % KE REMAINING & STATE THE TYPE OF COLLISION.
e] STATE THE ASSUMPTION(S).
10a A 10 g BALL MOVING AT 30 m/s COLLIDES WITH A 20 g BALL MOVING ALONG THE
SAME
LINE AT 10 m/s. AFTER COLLISION THE 10 g BALL IS MOVING ALONG THE SAME LINE
AT
15 m/s.
b] A 10 g BALL MOVING AT 30 m/s COLLIDES WITH A 20 g BALL MOVING TOWARD IT
AT
10 m/s.
AFTER COLLISION THE 10 g BALL IS MOVING IN THE OPPOSITE
DIRECTION AT 15 m/s.
PRACT: 20 g AT 40 m/s TO 25 m/s,
HITS 30 g AT 15 m/s TO ? m/s
a]
BEFORE
AFTER
Va = 30 m/s-----> Vb = 10
m/s--->
Va’ = 15 m/s-----> Vb’
= ? m/s--->
+---------+
+---------+
+---------+
+---------+
|Ma = 10 g|
|Mb = 20 g|
|Ma = 10 g|
|Mb = 20 g|
+---------+
+---------+
+---------+
+---------+
b] PREDICTION: Since Va lost speed
Vb must gain: Vb’ > 10 m/s
c] Ma = 10 g, Va = 30 m/s, Va’ = 15 m/s
Mb = 20 g, Vb = 10 m/s, Vb’
= ?
Eq: MaVa + MbVb
= MaVa’ + MbVb’
Sub: 10(30) + 20(10) = 10(15)
+ 20Vb’
Math: 300 + 200
= 150 + 20Vb’
500 = 150 + 20Vb’
-150 -150
.
350 = 20Vb’
Divide by 20--->Vb’ = 350/20 = 17.5 m/s
d]
MaVa'2 + MbVb'2
%K.E. REMAINING = ---------------100%
MaVa2 + MbVb2
10(152) + 20(17.52)
2250+6125 8,375
= ------------------100% =
---------100% = -------100% = 0.761x100%% = 76.1%
10(302) + 20(102)
9000+2000 11,000
76.1% Elastic 23.9% Heat,
Deformation, Sound
e] Assumptions: No external forces,
no friction, head on collisions.
10b]
A 10 g BALL MOVING AT 30 m/s COLLIDES WITH A 20 g BALL MOVING TOWARD IT AT
10 m/s.
AFTER COLLISION THE 10 g BALL IS MOVING IN THE OPPOSITE DIRECTION
AT 15 m/s.
a]
BEFORE
AFTER
Va = 30 m/s-----> Vb =
<---10 m/s Va’ =
<---15 m/s Vb’
= ? m/s--->
+---------+
+---------+
+---------+
+---------+
|Ma = 10 g|
|Mb = 20 g|
|Ma = 10 g|
|Mb = 20 g|
+---------+
+---------+
+---------+
+---------+
b] PREDICTION: Since Va lost speed
Vb must gain: Vb’ > 10 m/s
c] Ma = 10 g, Va = 30 m/s, Va’ = -15 m/s
Mb = 20 g, Vb = -10 m/s, Vb’
= ?
Eq: MaVa +
MbVb
= MaVa’ + MbVb’
Sub: 10(30) + 20(-10) =
10(-15)
+ 20Vb’
Math: 300 - 200
= -150 + 20Vb’
100 = -150 + 20Vb’
+150 +150
.
250 = 20Vb’
Divide by 20--->Vb’ = 250/20 = 12.5 m/s
d]
MaVa'2 + MbVb'2
%K.E. REMAINING = ---------------100%
MaVa2 + MbVb2
10(152) + 20(12.52)
2250+3125 5,375
= ------------------100% =
---------100% = -------100% = 0.4886x1000% = 48.9%
10(302) + 20(102)
9000+2000 11,000
48.9% Elastic
51.1% Heat,
Deformation, Sound
e] Assumptions: No external forces,
no friction, head on collisions.
PRACT: A 1,200 lb CANNON
FIRES A 53 lb SHELL AT 670 mi/h HORIZONTALLY.
11
TWO ASTRONAUTS, 80 kg AND 50 kg, FACE EACH OTHER AND PUSH
OFF. THE 80 kg ASTRONAUT MOVES AWAY AT 4 m/s.
a]
BEFORE
AFTER
Va = 0 m/s Vb
= 0 m/s
<----Va’ = 4 m/s
Vb’ = ? m/s--->
+----------+
+----------+
+----------+
+----------+
|Ma = 80 kg|
|Mb = 50 kg|
|Ma = 80 kg|
|Mb = 50 kg|
+----------+
+----------+
+----------+
+----------+
b] PREDICTION: Since Va + Vb =
0---->MaVa’ = -MbVb’
and Mb is lighter: Vb’
> Va, i.e. Vb’ > 4m/s
Since there is no
sticking--->Elastic Collision
c] Ma = 80 kg, Va = 0 m/s, Va’ = -4 m/s
Mb = 50 kg, Vb = 0 m/s, Vb’
= ?
Eq: MaVa + MbVb
= MaVa’
+ MbVb’
Sub: 80(0) + 50(0) = 80(-4) +
50Vb’
Math: 0 = -320
+ 50Vb’
+320 = +320
.
320 = 50Vb’
Divide by 50--->Vb’ = 320/50 = 6.4 m/s
Checks with prediction
d]
M1V1'2 + M2V2'2
M1V1'2 + M2V2'2
%K.E. REMAINING = --------------- = --------------- = INFINATE--->Not a
collision
M1V12 + M2V22
0
but a launch
e] Assumptions: No external forces,
no friction, head on collisions.
12 A 22 mi/h;
3000 lb TRUCK, COLLIDES WITH A 2000 lb CAR AT REST, THEY LOCK BUMPERS
PRACT: A 800 kg BOAT TRAVELS
AT 30 m/s AND COLLIDES WITH A 1600 kg TUG AT REST.
a]
BEFORE
AFTER
Vt= 22 mi/h---> Vc = 0 mi/h
Vt’ = Vc’ = V’ = ? mi/h--->
+-------------+ +-------------+
+-------------++-------------+
|Mt = 3,000 lb| |Mc
= 2,000 lb|
|Mt = 3,000 lb||Mc = 2,000 lb|
+-------------+ +-------------+
+-------------++-------------+
b] PREDICTION: Since start momentum
= final and the start momentum is in the truck
the added car mass will slow down the truck:
22 mi/h > V’ >0
Since there is
sticking--->Inelastic Collision
c] Mt = 3,000 lb, Vt = 22 mi/h, Sticking==>Vt’ = Vc’ = V’ =?
Mc = 2,000 lb, Vc =
0 mi/h
Eq: MtVt + McVc
= MtV’ + McV’ = (Mt +
Mc)V’
Sub: 3,000(22) +
2,000(0) = (3,000 +
2,000)V’
Math: 66,0000
=
5,000V’
Divide by 5,000--->V’ = 66,000/5,000 = 13.2 mi/h Checks with prediction
d]
MtVt'2 + McVc'2
(Mt + Mc)V’2
%K.E. REMAINING = ---------------100% = ---------------100%
MtVt2 + McVc2
MtVt2 + McVc2
(3000+2000)13.22
5,000x174.24 871,200
= ----------------100% = ------------100% = ---------100% = 60%
3000(222) + 0
1,452,000 1,452,000
60% Elastic 40% Heat,
Deformation, Sound
--->Doesn’t check with prediction!
even
though it sticks it is more than 50% elastic!
How elastic a collision is depends on the masses.
See algebraic analysis at the top.
e] Assumptions: No external forces,
no friction, head on collisions.
13
A 6 g BULLET STRIKES A WOODEN BLOCK OF MASS 500 g.
THE BULLET BECOMES EMBEDDED
IN THE BLOCK. THE
BLOCK WITH THE BULLET IN IT THEN FLIES OFF AT 12 m/s.
NEGLECT FRICTION & FIND THE BULLET'S VELOCITY.
a]
BEFORE
AFTER
Vb = ? m/s--> Vw = 0 m/s
Vb’ = Vw’ = V’ = 12 m/s--->
+--------+ +----------+
+------------------------+
|Mb = 6 g| |Mw
= 500 g|
|Mt = 6 g + Mw = 2,000 lb|
+--------+ +----------+
+------------------------+
b] PREDICTION: Since start momentum = final and the start momentum is in the
bullet
the
added wood mass will slow down the bullet:
Vb > 12 m/s
Since they
stick--->Inelastic Collision
c] Mb = 6 g, Vb = ?,
Sticking==>Vt’ = Vc’ = V’ = 12 m/s
Mw = 500 g, Vw =
0 m/s
Eq: MtVt + McVc =
MtV’ + McV’ = (Mt + Mc)V’
Sub: 6Vb + 500(0) =
(6 + 500)12 m/s
Math: 6Vb
= 6072 m/s
Divide by 5--->V’b = 6072/6 = 1012 m/s
Checks with prediction
d]
MbVb'2 + MwVw'2
(Mb + Mw)V’2
%K.E. REMAINING = ---------------100% = ---------------100%
MbVb2
+ MwVw2
MbVb2 + MwVw2
(6 + 500)122
72,864
= ------------100% = ---------100% = 1.19%
6(10122) + 0
6,144,864
1.19% Elastic 98.8% Heat,
Deformation, Sound
--->Checks with prediction!
14
Explain which will produce a greater momentum in object B:
a] If A hits B, then moves in the opposite direction; or
If A hits B, then stops.
Option 1: A hits B, then moves in the opposite direction
Pa = momentum of A
Before After
Collision Collision
+-+
+-+ +-+ +-+ Since
the original momentum is forward
|A|
|B| |A| |B| Pb has
to be forward to overcome the negative
+-+
+-+ +-+ +-+ momentum
of Pa',
---> &nbbsp; <---Pa' Pb'--->
Total Momentum = Total Momentum
Before After
Let Pa = 10 Pb = 0 Pa' =
-10 Pa + Pb = Paa' + Pb'
10 + 00 = -10 + Pb'--->Pb' = 20
Option 2: If A hits B, then stops.
Before After
Collision Collision
+-+
+-+ +-+ +-+
|A|
|B| |A| |B|
+-+
+-+ +-+ +-+ Total Momentum
= Total Momentum
---> &nbbsp; Pa' Pb'--->
Before After
Let Pa = 10 Pb = 0 Pa' =
0 Pa + Pb = Pa'+ Pb'
10 + 00 = 0 + Pb'--->Pb' = 10
Since
Pb' = 20 for option 1 > Pb' for option 2,
option 1 ( If A hits B, then moves in the opposite
direction) produced the greater momentum change in B.
b] If A hits B, sticks to B, then moves along with B in the
same direction; or
If A hits B, then stops?
Option 1: If A hits B, sticks to B, then moves along with B in the
same direction
Before After
Collision Collision
+-+
+-+ +-++-+
|A|
|B| |A||B|
+-+
+-+ +-++-+ Total Momentum =
Total Momentum
---> &nbbsp; Pa'Pb'--->
Before After
Let Pa = 10 Pb = 0 Pa'+Pb'=? 0 Pa + Pb
= Pa'+ Pb'
10 + 00 = Pa'+ Pb'
After Pa' & Pb' share the momentum of 10
Ex: Pa'= 8 & Pb'= 2: Note: 0<Pb'<10
Option 2: If A hits B, then stops.
Before After
Collision Collision
+-+
+-+ +-+ +-+
|A|
|B| |A| |B|
+-+
+-+ +-+ +-+ Total Momentum
= Total Momentum
---> &nbbsp; Pa' Pb'--->
Before After
Let Pa = 10 Pb = 0 Pa' =
0 Pa + Pb = Pa'+ Pb'
10 + 00 = 0 + Pb'--->Pb' = 10
Since Option 1 yields Pb'<10, and Option 2 yields Pb' = 10
Option 2 yields the greater change in B's momentum.
15
Explain which will produce a greater acceleration in object B:
a] If A hits B, then moves in the opposite direction; or
If A hits B, then stops.
Since the change in momentum = mΔV
and acceleration =
ΔV/t
they are both directly proportional to
ΔV.
That is, which ever option produces the greater change in momentum will also
produce the greater acceleration. The analysis above indicates that Option
1 will produce both the greater change in B's momentum and its
acceleration.
b] If A hits B, sticks to B, then moves along with B in the
same direction; or
If A hits B, then stops?
Similarly, the analysis above indicates that Option 2 will produce both the
greater change in B's momentum and its acceleration.
ANS:
4] 21.065 s;
5] 4.21 s,
34.1 lb, 2.05 s
6] 45.45 kg, 0.909 s; 150 lbs
10a] 17.5 m/s, 76.1% Elastic, 23.9%
Heat,
Deformation, Sound
10b]
12.5 m/s, 48.9% Elastic, 51.1% Heat, Deformation, Sound
11] -6.40 m/s, Not a collision but
a launch
12] 13.2 mph, , 60.0% Elastic, 40%
Heat
13] 1012 m/s,
1.19% Elastic, 98.8% HEAT
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