Download F - Uplift North Hills Prep

Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of motion for uniform
acceleration
EXAMPLE: How far will Pinky and the Brain go in 30.0
seconds if their acceleration is 20.0 m s -2?
KNOWN
FORMULAS
a = 20 m/s2 Given
s = ut + 12at2
t = 30 s
Given
v = u + at
Implicit
u = 0 m/s
v2 = u2 + 2as
s=?
WANTED
t is known - drop the
timeless eq’n.
Since v is not wanted,
drop the velocity eq'n:
SOLUTION
s = ut + 12at2
s = 0(30) + 12 20(30)2
s = 9000 m
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of motion for uniform
acceleration
EXAMPLE: How fast will Pinky and the Brain be going
at this instant?
KNOWN
a = 20 m/s2 Given
Given
t = 30 s
Implicit
u = 0 m/s
FORMULAS
s = ut + 12at2
v = u + at
v2 = u2 + 2as
v=?
WANTED
t is known - drop the
timeless eq’n.
Since v is wanted, drop
the displacement eq'n:
SOLUTION
v = u + at
v = 0 + 20(30)
v = 600 m s-1
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of motion for uniform
acceleration
EXAMPLE: How fast will Pinky and the Brain be going
when they have traveled a total of 18000 m?
KNOWN
a = 20 m/s2 Given
s = 18000 m Given
Implicit
u = 0 m/s
v=?
WANTED
Since t is not known drop the two eq’ns which
have time in them.
FORMULAS
s = ut + 12at2
v = u + at
v2 = u2 + 2as
SOLUTION
v2 = u2 + 2as
v2 = 02 + 2(20)(18000)
v = 850 m s-1
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of motion
for uniform acceleration
EXAMPLE: A ball is dropped off of the
Empire State Building (381 m tall). How fast
is it going when it hits ground?
KNOWN
FORMULAS
1 2
2
s
=
ut
+
a = -10 m/s Implicit
2at
v = u + at
s = -381 m Given
v2 = u2 + 2as
Implicit
u = 0 m/s
v=?
SOLUTION
WANTED
Since t is not
v2 = u2 + 2as
known - drop the
v2 = 02+ 2(-10)(-381)
two eq’ns which
v = -87 m s-1
have time in them.
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of motion
for uniform acceleration
EXAMPLE: A ball is dropped off of the
Empire State Building (381 m tall). How long
does it take to reach the ground?
KNOWN
a = -10 m/s2 Implicit
s = -381 m Given
Implicit
u = 0 m/s
FORMULAS
s = ut + 12at2
v = u + at
v2 = u2 + 2as
t=?
WANTED
Since t is desired
and we have s drop
the last two eq’ns.
SOLUTION
s = ut + 12at2
-381 = 0t + 12 (-10)t2
t = 8.7 s
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of
motion for uniform acceleration
EXAMPLE: A cheer leader is thrown up
with an initial speed of 7 m s-1. How high
does she go?
KNOWN
FORMULAS
s = ut + 12at2
a = -10 m/s2 Implicit
v = u + at
Given
u = 7 m s-1
v2 = u2 + 2as
Implicit
v = 0 m/s
s=?
WANTED
Since t is not known drop the two eq’ns which
have time in them.
SOLUTION
v2 = u2 + 2as
02 = 72 + 2(-10)s
s = 2.45 m
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of
motion for uniform acceleration
EXAMPLE: A ball is thrown upward at 50 m s-1 from the
top of the 300-m Millau Viaduct, the highest bridge in
the world. How fast does it hit ground?
KNOWN
FORMULAS
s = ut + 12at2
a = -10 m/s2 Implicit
v = u + at
u = 50 m s-1 Given
v2 = u2 + 2as
Implicit
s = -300 m
v=?
WANTED
Since t is not known drop the two eq’ns which
have time in them.
SOLUTION
v2 = u2 + 2as
v2 = 502 + 2(-10)(-300)
v = -90 m s-1
Topic 2: Mechanics
2.1 – Motion
Solving problems using equations of
motion for uniform acceleration
EXAMPLE: A ball is thrown upward at 50 m s-1 from the
top of the 300-m Millau Viaduct, the highest bridge in
the world. How long is it in flight?
KNOWN
FORMULAS
1 2
2
s
=
ut
+
a = -10 m/s
Implicit
2at
v = u + at
u = 50 m s-1
Given
v2 = u2 + 2as
v = -90 m s-1 Calculated
WANTED
Use the simplest t
equation.
t=?
SOLUTION
v = u + at
-90 = 50 + (-10)t
t = 14 s
Topic 2: Mechanics
2.1 – Motion
Sketching and interpreting motion graphs
EXAMPLE: Suppose Freddie the Fly begins at x = 0 m,
and travels at a constant velocity for 6 seconds as
shown. Find two points, sketch a displacement vs. time
graph, and then find and interpret the slope and the
area of your graph.
x/m t = 6 s, x = 18
t = 0, x = 0
SOLUTION:
The two points are (0 s, 0 m) and (6 s, 18 m).
The sketch is on the next slide.
Topic 2: Mechanics
2.1 – Motion
Sketching and interpreting motion graphs
x/m
SOLUTION:
27
24
21
18
15
12
9
6
3
0
Rise
s = 18 - 0
s = 18 m
t=6-0
Run t = 6 s
0
1
2
3
4
5
6
7
t/s
The slope is rise over run or 18 m / 6 s
Thus the slope is 3 m s-1, which is interpreted as
Freddie’s velocity.
8
9
Topic 2: Mechanics
2.1 – Motion
Sketching and interpreting motion graphs
VELOCITY (ms-1 )
EXAMPLE: Calculate and interpret the area under the
given v vs. t graph. Find and interpret the slope.
50
SOLUTION:
40
The area of a
30
20
triangle is
10
A = (1/2)bh.
0
10
0
5
Thus
TIME (sec)
A = (1/2)(20 s)(30 m/s) = 300 m.
This is the displacement of the object in 20 s.
The slope is (30 m/s) / 20 s = 1.5 m s-2.
This is the acceleration of the object.
15
20
t
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = uxt
∆y = uyt - 5t 2
vx = ux
vy = uy - 10t
reduced equations of
projectile motion
PRACTICE: A cannon fires a projectile with a muzzle
velocity of 56 ms-1 at an angle of inclination of 15º.
(a) What are ux and uy?
SOLUTION: Make a velocity triangle.
uy = u sin 
 = 15º
uy = 56 sin 15º
ux = u cos 
uy = 15 m s-1.
ux = 56 cos 15º
ux = 54 m s-1
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = uxt
∆y = uyt - 5t 2
vx = ux
vy = uy - 10t
reduced equations of
projectile motion
PRACTICE: A cannon fires a projectile with a muzzle
velocity of 56 ms-1 at an angle of inclination of 15º.
(b) What are the tailored equations of motion?
(c) When will the ball reach its maximum height?
SOLUTION: (b) Just substitute ux = 54 and uy = 15:
∆x = 54t
vx = 54
∆y = 15t - 5t2
vy = 15 - 10t
tailored equations for
this particular projectile
(c) At the maximum height, vy = 0. Why? Thus
vy = 15 - 10t becomes 0 = 15 - 10t so that
10t = 15 or t = 1.5 s.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = 54t
∆y = 15t - 5t 2
vx = 54
vy = 15 - 10t
tailored equations for
this particular projectile
PRACTICE: A cannon fires a projectile with a muzzle
velocity of 56 ms-1 at an angle of inclination of 15º.
(d) How far from the muzzle will the ball be when it
reaches the height of the muzzle at the end of its
trajectory?
SOLUTION:
From symmetry tup = tdown = 1.5 s so t = 3.0 s.
Thus
∆x = 54t
∆x = 54(3.0)
∆x = 160 m.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = 54t
∆y = 15t - 5t 2
vx = 54
vy = 15 - 10t
tailored equations for
this particular projectile
PRACTICE: A cannon fires a projectile with a muzzle
velocity of 56 ms-1 at an angle of inclination of 15º.
(e) Sketch the following graphs:
a vs. t,
vx vs. t,
vy vs. t:
SOLUTION: The only acceleration ay
t
-10
is g in the –y-direction.
vx
vx = 54, a constant. Thus it does 54
t
not change over time.
vy = 15 - 10t Thus it is linear with vy
a negative gradient and it crosses 15
t
1.5
the time axis at 1.5 s.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
The acceleration is ALWAYS g for projectile motionsince it is caused by Earth and its field.
At the maximum height the projectile switches from
upward to downward motion. vy = 0 at switch.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
The flight time is
limited by the y
motion.
The maximum
height is limited by
the y motion.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
ax = 0.
ay = -10 ms-2.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆y = uyt - 5t 2
-33 = 0t - 5t 2
-33 = -5t 2
(33/5) = t 2
Fall time limited by y-equations:
t = 2.6 s.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = uxt
∆x = 18(2.6)
Use x-equations and t = 2.6 s:
∆x = 47 m.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion

vx = ux
vx = 18.
vy = uy – 10t
vy = 0 – 10t
vy = –10(2.6) = -26.
tan  = 26/18
 = tan-1(26/18) = 55º.
18
26
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
The horizontal component of velocity is vx = ux which is
CONSTANT.
The vertical component of velocity is vy = uy – 10t ,
which is INCREASING (negatively).
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
 ∆EK + ∆EP = 0
∆EK = -∆EP
∆EK = -mg∆h
EKo = (1/2)mu2
∆EK = -(0.44)(9.8)(-32) = +138 J = EK – EKo
EK = +138 + (1/2)(0.44)(222) = 240 J.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
If 34% of the energy is
consumed, 76% remains.
0.76(240) = 180 J
(1/2)(0.44)v2 = 180 J
v = 29 ms-1.
(1/2)mvf2 - (1/2)mv2 = -∆EP
mvf2 = mv2 + -2mg(0-H)
Topic 2: Mechanics
2 = v2 + 2gH
v
f
2.1 – Motion
Analysing projectile motion
Use ∆EK + ∆EP = 0.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
uy = u sin 
uy = 28 sin 30º
ux = u cos 
ux = 28 cos 30º
ux = 24 m s-1.
uy = 14 m s-1.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
∆x = uxt
16 = 24t
t = 16 / 24 = 0.67
The time to the wall is found from ∆x…
∆y = uyt – 5t 2
∆y = 14t – 5t 2
∆y = 14(0.67) – 5(0.67)2 = 7.1 m.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
0.5s
0.0s
4m
ux = ∆x / ∆t = (4 - 0) / (0.5 - 0.0) = 8 ms-1.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
0.5s
11 m
0.0s
4m
uy = ∆y / ∆t = (11 - 0) / (0.5 - 0.0) = 22 ms-1.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
2.0 2.5 3.0
1.5
1.0
30 m
0.5s
11 m
0.0s

4m
 = tan-1(30/24) = 51º
24 m
D2 = 242 + 302 so that D = 38 m ,@  = 51º.
Topic 2: Mechanics
2.1 – Motion
Analysing projectile motion
New
peak
below and
left.
Pre-peak
greater
than postpeak.
Topic 2: Mechanics
2.2 – Forces
Objects as point particles and Free-body diagrams
where g = 10 m s -2
weight
W = mg
and m is the mass in kg
Free-body
diagram
mass
force
EXAMPLE: Calculate the weight of a 25-kg
object.
SOLUTION:
Since m = 25 kg and g = 10 m s-2,
W = mg = (25)(10) = 250 N (or 250 n).
Note that W inherits its direction from the fact
that g points downward.
We sketch the mass as a point particle (dot),
and the weight as a vector in a free-body
diagram:
W
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
The resultant (or net) force is just the vector sum of
all of the forces acting on a body.
EXAMPLE: An object has mass of 25 kg. A tension of
50 n and a friction force of 30 n are acting on it as
shown. What is the resultant force?
SOLUTION:
R
Since the weight and the normal
forces cancel out in the y-direction,
50 n
we only need to worry about the
Ff
forces in the x-direction.
30 n
The net force is thus
W
50 – 30 = 20 n (+x-dir).
T
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
The resultant (or net) force is just the vector sum of
all of the forces acting on a body.
Fx,net = Fx
Fnet = F
Fy,net = Fy
net force
30. n
EXAMPLE: An object has exactly two forces F1 = 50. n
and F2 = 30. n applied simultaneously to it. What is the
resultant force’s magnitude?
F2
SOLUTION:
Fnet = F = F1 + F2 so we simply
graphically add the two vectors:
The magnitude is given by
50. n F1
Fnet2 = 502 + 302
Fnet = 58 n.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
The resultant (or net) force is just the vector sum of
all of the forces acting on a body.
Fx,net = Fx
Fnet = F
Fy,net = Fy
net force
F2
30. n
EXAMPLE: An object has exactly two forces F1 = 50. n
and F2 = 30. n applied simultaneously to it as shown.
What is the resultant force’s direction?
SOLUTION:
Direction is measured from the (+) x-axis.
Opposite and adjacent are given directly,

so use tangent.
50. n F1
tan  = opp / adj = 30 / 50 = 0.6
 = tan-1(0.6) = 31°.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
Fnet,y = 23 + 30 = 53 n.
Fnet2 = Fnet,x2 + Fnet,y2
Fnet2 = 442 + 532
Fnet = 69 n.
28°
44 n
50 cos 28
23 n
50 sin 28
Then Fnet,x = 44 n and
30. n
EXAMPLE: An object has exactly two forces F1 = 50. n
and F2 = 30. n applied simultaneously to it. What is the
resultant force’s magnitude?
SOLUTION:
Begin by resolving F1 into its xF2
and y-components.
Topic 2: Mechanics
2.2 – Forces
y
R
Describing solid friction by coefficients of friction
Ff
x
15°
mg
EXAMPLE: A piece of wood with a coin on it is
FBD, coin
raised on one end until the coin just begins to
slip. The angle the wood makes with the
horizontal is θ = 15°. What is the
coefficient of static friction?
θ = 15°
∑Fy = 0
R – mg cos 15° = 0
R = mg cos 15°
∑Fx = 0
Ff – mg sin 15° = 0
Ff = mg sin 15°
Ff = μs N
μs = mg sin 15° = tan 15°
mg cos 15°
mg sin 15° = μs mg cos 15°
= 0.268
Thus the coefficient of static friction between the metal
of the coin and the wood of the plank is 0.268.
Topic 2: Mechanics
2.2 – Forces
Describing solid friction by coefficients of friction
y
R
Ff
x
12°
mg
EXAMPLE: Now suppose the plank of wood is
long enough so that you can lower it to the point FBD, coin
that the coin keeps slipping, but no longer accelerates
(v = 0). If this new angle is 12°, what is the coefficient of
dynamic friction?
θ = 12°
∑Fx = 0
∑Fy = 0
Ff – mg sin 12° = 0
R – mg cos 12° = 0
Ff = mg sin 12°
R = mg cos 12°
Fd = μd R
 μd = tan 12° = 0.213
mg sin 12° = μd mg cos 12°
Thus the coefficient of dynamic friction between the
metal of the coin and the wood of the plank is 0.213.
30° 45° T2
Topic 2: Mechanics
2.2 – Forces
T1
T3
m
Translational equilibrium
EXAMPLE: An object of mass m is hanging via
T3
three cords as shown. Find the tension in each
of the three cords, in terms of m.
SOLUTION:
T3 = mg
mg
Now we break T1 and T2 down to components.
FBD, m
Looking at the FBD of the knot we see that
T1x = T1 cos 30° = 0.866T1
T2
T1
T1y = T1 sin 30° = 0.500T1
30° 45°
T2x = T2 cos 45° = 0.707T2
T2y = T2 sin 45° = 0.707T2
T
3
FBD, knot
30° 45° T2
Topic 2: Mechanics
2.2 – Forces
T1
T3
m
Translational equilibrium
EXAMPLE: An object of mass m is hanging via
T3
three cords as shown. Find the tension in each
of the three cords, in terms of m.
SOLUTION:
T3 = mg
∑Fx = 0
mg
0.707T2 - 0.866T1 = 0
FBD, m
T2 = 1.225T1
∑Fy = 0
T2
T1
0.707T2 + 0.500T1 - T3 = 0
0.707(1.225T1) + 0.500T1 = T3
30° 45°
T1 = mg / 1.366
T3
T2 = 1.225(mg / 1.366)
FBD, knot
T2 = 0.897mg
30° 45° T2
Topic 2: Mechanics
2.2 – Forces
T1
T3
m
Solving problems involving forces and resultant force
PRACTICE: A 25-kg mass is hanging via three cords as
shown. Find the tension in each of the three cords, in
Newtons.
SOLUTION:
Since all of the angles are the same use the formulas
we just derived:
T3 = mg = 25(10) = 250 n
T1 = mg / 1.366 = 25(10) / 1.366 = 180 n
T2 = 0.897mg = 0.897(25)(10) = 220 n
FYI This was an example of using Newton’s first law
with v = 0. The next example shows how to use
Newton’s first law when v is constant, but not zero.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
EXAMPLE: A 1000-kg airplane is flying at a constant
velocity of 125 m s-1. Label and determine the value of
the weight W, the lift L, the drag D and the thrust F if the
drag is 25000 N.
L
SOLUTION: D
W
Since the velocity is constant,
Newton’s first law applies. Thus Fx = 0 and Fy = 0.
W = mg = 1000(10) = 10000 N (down).
Since Fy = 0, L - W = 0, so L = W = 10000 N (up).
D = 25000 N tries to impede the aircraft (left).
Since F = 0, F - D = 0, so F = D = 25000 N (right).
F
Topic 2: Mechanics
2.2 – Forces
Newton’s laws of motion – The second law
Fnet = ma (or F = ma )
Newton’s second law
EXAMPLE: An object has a mass of 25 kg. A tension of
R
50 n and a friction force of 30 n are acting
on it as shown. What is its acceleration?
50 n
SOLUTION:
Ff
The vertical forces W and R
30 n
cancel out.
W
The net force is thus
Fnet = 50 – 30 = 20 n (+x-dir).
From Fnet = ma we get 20 = 25 a so that
a = 20 / 25 = 0.8 m s-2 (+x-dir).
T
Topic 2: Mechanics
2.2 – Forces
Newton’s laws of motion – The second law
Fnet = ma (or F = ma )
Newton’s second law
PRACTICE: Use F = ma to show that the formula for
weight is correct.
SOLUTION:
F = ma.
But F is the weight W.
And a is the freefall acceleration g.
Thus F = ma becomes W = mg.
Topic 2: Mechanics
2.2 – Forces
Newton’s laws of motion – The second law
Fnet = ma (or F = ma )
Newton’s second law
EXAMPLE: A 1000-kg airplane is flying in perfectly level
flight. The drag D is 25000 n and the thrust F is 40000
n. Find its acceleration.
L
D
SOLUTION:
W
Since the flight is level, Fy = 0.
Fx = F – D = 40000 – 25000 = 15000 n = Fnet.
From Fnet = ma we get 15000 = 1000a, or
a = 15000 / 1000 = 15 m s-2.
F
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
17 n
40 sin 25
30 n
EXAMPLE: A 25-kg object has exactly two forces F1 =
40. n and F2 = 30. n applied simultaneously to it. What
F2
is the object’s acceleration?
SOLUTION:
Resolve F1 into its components:
Then Fnet,x = 36 n and
25°
Fnet,y = 17 + 30 = 47 n. Then
36 n
Fnet2 = Fnet,x2 + Fnet,y2
40 cos 25
Fnet2 = 362 + 472 and Fnet = 59 n.
Then from Fnet = ma we get 59 = 25a, or
a = 59 / 25 = 2.4 m s-2.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
6.0 m
EXAMPLE: A 25-kg object resting
R
on a frictionless incline is released,
as shown. What is its acceleration?
60
SOLUTION:
30°
mg
cos
30
Begin with a FBD.
mg sin 30
mg
Break down the weight into its components.
Since R and mg cos 30°are perpendicular to the path
of the crate they do NOT contribute to its acceleration.
Thus
Fnet = ma
mg sin 30° = ma
a = 10 sin 30° = 5.0 m s-2.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
6.0 m
u=0
EXAMPLE: A 25-kg object resting
on a frictionless incline is released,
as shown. What is its speed at the
bottom?
v=?
30°
SOLUTION:
We found that its acceleration is 5.0 m s-2.
We will use v 2 = u 2 + 2as to find v, so we need s.
We have opposite and we want hypotenuse s so from
trigonometry, we use
sin  = opp / hyp.
Thus s = hyp = opp / sin  = 6 / sin 30° = 12 m and
v2 = u2 + 2as = 02 + 2(5)(12) = 120
so that v = 11 m s-1.
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
EXAMPLE: A 100.-n crate is to be
y
R F
dragged across the floor by an applied
force F = 60 n, as shown. The
Ff
30°
x
coefficients of static and dynamic friction
a
are 0.75 and 0.60, respectively. What is
mg
the acceleration of the crate?
FBD, crate
SOLUTION:
Static friction will oppose the applied force until it is
overcome.
F
FYI Since friction is proportional to
N
30° the normal force, be aware of
a problems where an applied force
Ff
changes the normal force.
mg
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
SOLUTION:
y
R F
Determine if the crate even moves.
Ff
Thus, find the maximum value of the
30°
x
static friction, and compare it to the
a
horizontal applied force:
mg
FH = F cos 30° = 60 cos 30° = 51.96 n.
FBD, crate
The maximum static friction force is
Fs,max = μs R = 0.75R
The normal force is found from...
R + F sin 30° - mg = 0
R + 60 sin 30° - 100 = 0 R = 70
Fs,max = 0.75(70) = 52.5 N
Thus the crate will not even begin to move!
Topic 2: Mechanics
2.2 – Forces
Solving problems involving forces and resultant force
EXAMPLE: If someone gives the crate a
small push (of how much?) it will “break”
loose. What will its acceleration be then?
SOLUTION:
The horizontal applied force is still
F cos 30° = 60 cos 30° = 51.96 n.
The dynamic friction force is
Fd = μd R = 0.60R.
The reaction force is still R = 70. n.
Thus Fd = 0.60(70) = 42 n.
The crate will accelerate.
F cos 30° - Fd = ma
51.96 - 42 = (100 / 10)a
a = 0.996 m/s2
y
Ff
R
F
30°
mg
FBD, crate
x
a
Topic 2: Mechanics
2.2 – Forces
Identifying force pairs in context of Newton’s third law
EXAMPLE: When you push on a door
with 10 n, the door pushes you back
with exactly the same 10 n, but in the
opposite direction. Why does the door
move, and not you?
FBA
FAB
SOLUTION: Even though the forces
are equal and opposite, they are
B
acting on different bodies.
A A
Each body acts in response only to the force
acting on it.
The door CAN’T resist FAB, but you CAN resist FBA.
Topic 2: Mechanics
2.2 – Forces
FBE
Identifying force pairs in context of Newton’s third law
NBT
EXAMPLE:
Consider a baseball resting on a
NTB
tabletop. Discuss each of the forces
acting on the baseball, and the
associated reaction force.
SOLUTION:
FEB
Acting on the ball is its weight FBE
prior to contact with the table.
Note that FBE (the weight force) and NBT (the normal
force) are acting on the ball.
NTB (the normal force) acts on the table.
FEB (the weight force) acts on the earth.
Topic 2: Mechanics
2.3 – Work, energy, and power
Determining work done by a force
EXAMPLE: Suppose we wish to find the speed of the
ball when it reaches the bottom of the track. Discuss the
problems in using free-body diagrams to find that final
speed.
W N
KEY
SOLUTION: Because the slope of the track is changing,
so is the relative orientation of N and W.
Thus, the acceleration is not constant and we can’t use
the kinematic equations.
Thus we can’t find v at the bottom of the track.
Topic 2: Mechanics
2.3 – Work, energy, and power
Determining work done by a force
W = Fs
work done by a constant force
EXAMPLE: Find the work done by the 25-Newton force
F in displacing the box s = 15 meters.
s
F
SOLUTION:
W = Fs
W = (25 N)(15 m)
W = 380 N m = 380 J.
FYI The units of (N m) are Joules (J). You can just
keep them as (N m) if you prefer.
Topic 2: Mechanics
2.3 – Work, energy, and power
Determining work done by a force
W = Fs cos 
work done by a constant force
not parallel to displacement
Where  is the angle between F and s.
PRACTICE: Find the work done by the force F = 25 N in
displacing a box s = 15 m if the force and displacement
are (a) parallel, (b) antiparallel and (c) at a 30° angle.
SOLUTION:
(a)
W = Fs cos  = (25)(15) cos 0° = 380 J.
(b)
W = (25)(15) cos 180° = - 380 J.
(c)
W = (25)(15) cos 30° = 320 J.
FYI Work can be negative.
F and the s are the magnitudes of F and s.
Topic 2: Mechanics
2.3 – Work, energy, and power
Determining work done by a force
EXAMPLE: Find the work done by the brakes in
bringing a 730-kg Smart Car to a rest in 80. meters if its
starting speed is 32 m/s.
F
s
SOLUTION: F and s are antiparallel so  = 180°.
From s = 80 m and v2 = u2 + 2as we get
02 = 322 + 2a(80) so that a = -6.4 m s-2.
Then F = ma = 730(-6.4) = - 4672 n. |F| = +4672 N.
Finally, W = Fs cos 
= (4672)(80) cos 180° = - 370000 J.
FBD Crate
Topic 2: Mechanics
2.3 – Work, energy, and power
Determining work done by a force
EXAMPLE: A pulley system is used to raise a
100-N crate 4 m as shown. Find the work done
by the tension force T if the lift occurs at constant
speed.
SOLUTION:
From the FBD since a = 0, T = 100 N.
From the statement of the problem, s = 4 m.
Since the displacement and the tension are
parallel,  = 0°.
Thus W = Ts cos  = (100)(4) cos 0° = 400 J.
FYI
Pulleys are used to redirect tension forces.
T
a=0
100
T
sT
s
T
T
FBD Crate
Topic 2: Mechanics
2.3 – Work, energy, and power
T
T
a=0
Determining work done by a force
100
EXAMPLE: A pulley system is used to raise a
100-N crate 4 m as shown. Find the work done
by the tension force T if the lift occurs at constant
speed.
T
T
SOLUTION:
From the FBD 2T = 100 so that T = 50 n.
From the statement of the problem, s = 4 m.
T 2s
Since the displacement and the tension are
T
parallel,  = 0°.
s
T
So W = T(2s) cos  = (50)(24) cos 0° = 400 J.
FYI
M.A. = Fout / Fin = 100 / 50 = 2.
Pulleys are also used gain mechanical advantage.
Topic 2: Mechanics
2.3 – Work, energy, and power
Sketching and interpreting force – distance graphs
F = - ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a spring is
shown. Find the value of the spring constant, and find
the spring force if the displacement is -65 mm.
F/N
SOLUTION:
20
Pick any convenient point.
s/mm
For this point F = -15 N and
0
s = 30 mm = 0.030 m so that
F = -ks or -15 = -k(0.030) 20
-40
-20
-1
20
0
40
k = 500 N m .
F = -ks = -(500)(-6510-3) = +32.5 n.
Topic 2: Mechanics
2.3 – Work, energy, and power
Sketching and interpreting force – distance graphs
F = - ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a spring is
shown. Find the work done by you if you displace the
spring from 0 to 40 mm.
SOLUTION:
F/N
20
The graph shows the force F
of the spring, not your force.
s/mm
0
The force you apply will be
opposite to the spring’s force
-20
according to F = +ks.
-40
-20
20
0
40
+
F = ks is plotted in red.
Topic 2: Mechanics
2.3 – Work, energy, and power
Sketching and interpreting force – distance graphs
F = - ks
Hooke’s Law (the spring force)
EXAMPLE: A force vs. displacement plot for a spring is
shown. Find the work done by you if you displace the
spring from 0 to 40 mm.
F/N
SOLUTION:
20
The area under the F vs. s
graph represents the work
s/mm
0
done by that force.
The area desired is from 0 mm
-20
to 40 mm, shown here:
-40
-20
20
0
40
A = (1/2)bh = (1/2)(4010-3 m)(20 N) = 0.4 J.
Topic 2: Mechanics
2.3 – Work, energy, and power
Elastic potential energy
EP = (1/2)kx 2
Elastic potential energy
EXAMPLE: Show that the energy F
“stored” in a stretched or compressed
spring is given by the above formula.
SOLUTION:
We equate the work W done in
deforming a spring (having a spring constant k by a
displacement x) to the energy EP “stored” in the spring.
If the deformed spring is released, it will go back to its
“relaxed” dimension, releasing all of its stored-up
energy. This is why EP is called potential energy.
s
Topic 2: Mechanics
2.3 – Work, energy, and power
Elastic potential energy
EP = (1/2)kx 2
Elastic potential energy
EXAMPLE: Show that the energy F
“stored” in a stretched or compressed
spring is given by the above formula.
SOLUTION:
As we learned, the area under the
F vs. s graph gives the work done by the force during
that displacement.
From F = ks and from A = (1/2)bh we obtain
EP = W = A = (1/2)sF = (1/2)s×ks = (1/2)ks2.
Finally, since s = x, EP = (1/2)kx2.
s
Topic 2: Mechanics
2.3 – Work, energy, and power
Kinetic energy
EK = (1/2)mv 2
PRACTICE: What is the kinetic
energy of a 4.0-gram NATO SS
109 bullet traveling at 950 m/s?
SOLUTION:
Convert grams to kg (jump 3
decimal places to the left) to get
m = 0.004 kg.
Then EK = (1/2)mv 2
= (1/2)(.004)(950) 2
= 1800 J.
kinetic energy
Topic 2: Mechanics
2.3 – Work, energy, and power
Kinetic energy
EK = (1/2)mv 2
kinetic energy
EXAMPLE: What is the kinetic
energy of a 220-pound NATO
soldier running at 6 m/s?
SOLUTION:
First convert pounds to kg:
(220 lb)(1 kg / 2.2 lb) = 100 kg.
Then EK = (1/2)mv 2
= (1/2)(100)(6) 2 = 1800 J.
FYI
Small and large objects can have the same EK!
Topic 2: Mechanics
2.3 – Work, energy, and power
Work done as energy transfer
W = ∆EK
work-kinetic energy theorem
EXAMPLE: Use energy to find the work done by the
brakes in bringing a 730-kg Smart Car to a rest in 80.
meters if its starting speed is 32 m/s.
F
s
SOLUTION:
EK,f = (1/2)mv 2 = (1/2)(730)(02) = 0 J.
EK,0 = (1/2)mu 2 = (1/2)(730)(322) = 370000 J.
∆EK = EK,f - EK,0 = 0 – 370000 = - 370000 J.
W = ∆E = -370000 J (same as before, easier!)
Topic 2: Mechanics
2.3 – Work, energy, and power
Gravitational potential energy
∆EP = mg∆h
gravitational potential energy change
PRACTICE: Consider a crane which lifts a
2000-kg weight 18 m above its original
resting place. What is the change in
gravitational potential energy of the weight?
SOLUTION:
The change in gravitational potential
energy is just
∆EP = mg∆h = 2000(10)(18) = 360000 J.
FYI Note that the units for ∆EP are those
of both work and kinetic energy.
Topic 2: Mechanics
2.3 – Work, energy, and power
Work done as energy transfer
W = Fs
work done by a constant force
PRACTICE: Consider a crane which lifts a
2000-kg weight 18 m above its original
resting place. How much work does the
crane do?
SOLUTION:
The force F = mg = 2000(10) = 20000 N .
The displacement s = 18 m .
Then W = Fs = 20000(18) = 360000 J.
FYI Note that the work done by the crane
is equal to the change in potential energy..
Topic 2: Mechanics
2.3 – Work, energy, and power
Principle of conservation of energy
EK = (1/2)mv 2
kinetic energy
EXAMPLE: Consider a crane which lifts a
2000-kg weight 18 m above its original
resting place. If the cable breaks at the top,
find the speed and kinetic energy of the
mass at the instant it reaches the ground.
SOLUTION: a = -g because it is freefalling.
v2 = u2 + 2as
v2 = 02 + 2(-10)(-18) = 360
v = 18.97366596 m s -1.
EK = (1/2)mv 2
= (1/2)(2000)(18.97367 2) = 360000 J.
Topic 2: Mechanics
2.3 – Work, energy, and power
Principle of conservation of energy
∆EP = mg∆h
gravitational potential energy change
EXAMPLE: Consider a crane which lifts a
2000-kg weight 18 m above its original
resting place. If the cable breaks at the top
find its change in kinetic energy and
change in potential energy the instant it
reaches the ground.
SOLUTION: EK,0 = (1/2)(2000)(02) = 0 J.
EK,f = 360000 J (from last slide).
∆EK = 360000 – 0 = 360000 J.
∆EP = mg∆h
= (2000)(10)(-18) = -360000 J.
Topic 2: Mechanics
2.3 – Work, energy, and power
Principle of conservation of energy
∆EP = mg∆h
gravitational potential energy change
EXAMPLE: Consider a crane which lifts a
2000-kg weight 18 m above its original
resting place. If the cable breaks at the top
find the sum of the change in kinetic and
the change in potential energies the instant
it reaches the ground.
SOLUTION: From the previous slide
∆EK = 360000 J and
∆EP = -360000 J so that
∆EK + ∆EP = 360000 + - 360000 = 0.
Hence ∆EK + ∆EP = 0 J.
Topic 2: Mechanics
2.3 – Work, energy, and power
Discussing the conservation of total energy
∆EK + ∆EP = 0
conservation of energy
In the absence of friction and drag
EXAMPLE: Find the speed of the 2-kg ball when it
reaches the bottom of the 20-m tall frictionless track.
∆h
SOLUTION: Use energy conservation to find EK,f and v.
∆EK + ∆EP = 0 FYI If friction is
(1/2)mv2 - (1/2)mu2 + mg∆h = 0 zero, m always
(1/2)(2)v2 - (1/2)(2)02 + 2(10)(-20) = 0 cancels…
v2 = 400  v = 20 m s-1.
Topic 2: Mechanics
2.3 – Work, energy, and power
Discussing the conservation of total energy
6.0 m
EXAMPLE: A 25-kg object resting
u=0
on a frictionless incline is
released, as shown. What is its
∆h
speed at the bottom?
30°
SOLUTION: We solved this one
long ago using Newton’s second law. It was difficult!
We will now use energy to solve it.
∆EK + ∆EP = 0
(1/2)mv 2 - (1/2)mu 2 + mg∆h = 0
(1/2)(25)v 2 - (1/2)(25)0 2 + (25)(10)(-6) = 0
12.5v 2 = 1500
FYI If friction and drag are
zero, m always cancels…
v = 11 m s-1.
v=?
Topic 2: Mechanics
2.3 – Work, energy, and power
Discussing the conservation of total energy within
energy transformations
EXAMPLE: Suppose the speed of the 2-kg ball is 15 m
s-1 when it reaches the bottom of the 20-m tall track.
Find the loss of mechanical energy and its “location.”
The system lost 175 J as drag and friction heat.
∆h
SOLUTION: Use ∆EK + ∆EP = loss or gain.
(1/2)mv2 - (1/2)mu2 + mg∆h = loss or gain
(1/2)(2)152 - (1/2)(2)02 + 2(10)(-20) = loss or gain
- 175 J = loss or gain
Topic 2: Mechanics
2.3 – Work, energy, and power
Discussing the conservation of total energy within
energy transformations
EK + EP = ET = CONST
EXAMPLE: Suppose the simple
pendulum shown has a 1.25-kg
“bob” connected to a string that
is 0.475 m long. Find the maximum FYI Assume the
velocity of the bob during its cycle. drag force is zero.
SOLUTION: Use ∆EK + ∆EP = 0.
Maximum kinetic energy occurs at the lowest point.
Maximum potential energy occurs at the highest point.
(1/2)mv 2 - (1/2)mu 2 + mg∆h = 0
(1/2)(1.25)v 2 - (1/2)(1.25)0 2 + 1.25(10)(-0.475) = 0
v = 3.08 ms-1.
Topic 2: Mechanics
2.3 – Work, energy, and power
Discussing the conservation of total energy within
energy transformations
EK + EP = ET = CONST
EXAMPLE: Suppose a
1.25-kg mass is
connected to a spring that
x
-1
has a constant of 25.0 Nm and
FYI Assume the
is displaced 4.00 m before being
friction force is zero.
released. Find the maximum
velocity of the mass during its cycle.
SOLUTION: Use ∆EK + ∆EP = 0 and v = vmax at x = 0.
(1/2)mv 2 - (1/2)mu 2 + (1/2)k∆xf2 – (1/2)k∆x02 = 0
(1/2)(1.25)v 2 + (1/2)(25)0 2 – (1/2)(25)(42) = 0
v = 17.9 ms-1.
Topic 2: Mechanics
2.3 – Work, energy, and power
Power as rate of energy transfer
Power is the rate of energy usage and so has the
equation
P=E/t
power
From the formula we see that power has the units of
energy (J) per time (s) or (J s-1) which are known as
watts (W).
EXAMPLE: How much energy does a 100.-W bulb
consume in one day?
SOLUTION: From P = E / t we get E = Pt so that
E = (100 J/s)(24 h)(3600 s/h)
E = 8640000 J!
Don’t leave lights on in unoccupied rooms.
Topic 2: Mechanics
2.3 – Work, energy, and power
Power as rate of energy transfer
P = Fv cos 
power
PRACTICE: Show that P = Fv cos .
SOLUTION: Since P = E / t we can begin by rewriting
the energy E as work W = Fs cos  :
P=E/t
=W/t
= Fs cos  / t
= F (s / t) cos 
= Fv cos .
FYI
The Physics Data Booklet has only “P = Fv.”
Topic 2: Mechanics
2.3 – Work, energy, and power
Power as rate of energy transfer
P = Fv
the last horse-drawn
barge operated on the
River Lea ...(1955)
power
EXAMPLE: Sam the horse, walking
at 1.75 ms-1, is drawing a barge
v

F
having a drag force of 493 N along
the River Lea as shown. The angle
the draw rope makes with the
velocity of the barge is 30. Find the
rate at which Sam is expending energy.
SOLUTION: Since energy rate is power, use
P = Fv cos 
FYI Since 1 horsepower is
= (493)(1.75) cos 30 746 W, Sam is earning his
keep, exactly as planned!
= 747 W.
Topic 2: Mechanics
2.3 – Work, energy, and power
Power as rate of energy transfer
P = Fv
power
EXAMPLE: The drag force of a moving object is
approximately proportional to the square of the velocity.
Find the ratio of the energy rate of a car traveling at 50
mph, to that of the same car traveling at 25 mph.
SOLUTION: Since energy rate is power, use P = Fv.
Then F = Kv 2 for some K and P = Fv = Kv 2v = Kv 3.
Thus
FYI
P50 / P25 = K503 / K253 It takes 8 times as much
= (50 / 25)3
gas just to overcome air
resistance if you double
= 23
your speed! Ouch!
= 8.
Topic 2: Mechanics
2.3 – Work, energy, and power
Quantitatively describing efficiency in energy transfers
Efficiency is the ratio of output power to input power
efficiency = Wout / Win = Pout / Pin
efficiency
EXAMPLE: Conversion of coal into electricity is through
the following process: Coal burns to heat up water to
steam. Steam turns a turbine. The turbine turns a
generator which produces electricity. Suppose the
useable electricity from such a power plant is 125 MW,
while the chemical energy of the coal is 690 MW. Find
the efficiency of the plant.
SOLUTION:
efficiency = Pout / Pin
= 125 MW / 690 MW
= 0.18 or 18%.
Topic 2: Mechanics
2.4 – Momentum and impulse
Newton’s second law in terms of momentum
Linear momentum, p, is defined to be the product of
an object’s mass m with its velocity v.
p = mv
linear momentum
Its units are obtained directly from the formula and are
kg m s-1.
EXAMPLE: What is the linear momentum
of a 4.0-gram NATO SS 109 bullet
traveling at 950 m/s?
SOLUTION:
Convert grams to kg (jump 3 decimal
places left) to get m = .004 kg.
Then p = mv = (.004)(950) = 3.8 kg m s-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Newton’s second law in terms of momentum
p = mv
linear momentum
Fnet = ma = m (v / t ) = ( m v ) / t = p / t.
This last is just Newton’s second law in terms of
change in momentum rather than mass and
acceleration.
Fnet = p / t
Newton’s second law (p-form)
EXAMPLE: A 6-kg object increases its speed from 5 m
s-1 to 25 m s-1 in 30 s. What is the net force acting on it?
SOLUTION:
Fnet = p / t = m( v – u ) / t
= 6( 25 – 5 ) / 30 = 4 N.
Topic 2: Mechanics
2.4 – Momentum and impulse
Kinetic energy in terms of momentum
p = mv
linear momentum
EK = (1/2)mv 2
kinetic energy
EXAMPLE: Show that kinetic energy can be calculated
directly from the momentum using the following:
EK = p 2 / (2m)
kinetic energy
SOLUTION:
From p = mv we obtain v = p / m. Then
EK = (1/2) mv 2
= (1/2) m (p / m)2
= mp 2 / (2m2)
= p 2 / (2m)
Topic 2: Mechanics
2.4 – Momentum and impulse
Kinetic energy in terms of momentum
EK = p 2 / (2m)
PRACTICE: What is the kinetic
energy of a 4.0-gram NATO SS 109
bullet traveling at 950 m/s and having
a momentum of 3.8 kg m s-1?
SOLUTION: You can work from
scratch using EK = (1/2)mv 2 or you
can use EK = p 2 / (2m).
Let’s use the new formula…
EK = p 2 / (2m)
= 3.8 2 / (2×0.004)
= 1800 J.
kinetic energy
Topic 2: Mechanics
2.4 – Momentum and impulse
Impulse and force – time graphs
EXAMPLE: A 0.140-kg baseball comes in at 40.0 m/s,
strikes the bat, and goes back out at 50.0 m/s. If the
collision lasts 1.20 ms (a typical value), find the average
force exerted on the ball during the collision.
SOLUTION: We can use J = F t. Thus
Fmax
F = J / t
= 12.6 / 1.20×10-3
F
= 10500 N.
FYI
Since a Newton is about a quarter-pound, F is about
10500 / 4 = 2626 pounds – more than a ton of force!
Furthermore, Fmax is even greater than F!
Topic 2: Mechanics
2.4 – Momentum and impulse
Sketching and interpreting force – time graphs
J = F ∆t = p = area under F vs. t graph
impulse
Force F / n
PRACTICE: A bat striking a ball imparts a force to it as
shown in the graph. Find the impulse.
SOLUTION:
Break the graph into simple areas of rectangles and
triangles.
9
 A1 = (1/2)(3)(9) = 13.5 N s
6
 A2 = (4)(9) = 36 N s
3
 A3 = (1/2)(3)(9) = 13.5 N s
0
Atot = A1 + A2 + A3
0
5
10
Time t / s
Atot = 13.5 + 36 + 13.5 = 63 N s.
Topic 2: Mechanics
2.4 – Momentum and impulse
Impulse and force – time graphs
EXAMPLE:
T
v
How does a jet engine
produce thrust?
SOLUTION:
The jet engine sucks
in air (at about the speed
that the plane is flying through the air), heats it up, and
expels it at a greater velocity.
The momentum of the air changes since its velocity
does, and hence an impulse has been imparted to it by
the engine.
The engine feels an equal and opposite impulse.
Hence the engine creates a thrust.
Topic 2: Mechanics
2.4 – Momentum and impulse
This is a 2stage rocket.
The orange
tanks hold fuel,
and the blue
tanks hold
oxidizer.
The oxidizer is
needed so that
the rocket works
without air.
Impulse and force – time graphs
EXAMPLE:
Show that F = (∆m / ∆t )v.
SOLUTION:
From F = p / t we have
F = p / t
F = (mv) / t
F = ( m / t )v (if v is constant).
FYI
The equation F = ( ∆m / ∆t )v is known as the rocket
engine equation because it shows us how to calculate
the thrust of a rocket engine.
The second example will show how this is done.
Topic 2: Mechanics
2.4 – Momentum and impulse
Impulse and force – time graphs
T
EXAMPLE:
What is the purpose of the rocket nozzle?
SOLUTION:
In the combustion chamber the gas
particles have random directions.
The shape of the nozzle is such that
the particles in the sphere of combustion
are deflected in such a way that they all
come out antiparallel to the rocket.
This maximizes the impulse on the gases.
The rocket feels an equal and opposite (maximized)
impulse, creating a maximized thrust.
Topic 2: Mechanics
2.4 – Momentum and impulse
Impulse and force – time graphs
F = ( m / t )v
rocket engine equation
EXAMPLE: A rocket engine consumes
fuel and oxidizer at a rate of 275 kg s-1
and used a chemical reaction that gives
the product gas particles an average
speed of 1250 ms-1. Find the thrust
produced by this engine.
SOLUTION:
The units of m / t are kg s-1 so that
clearly m / t = 275.
The speed v = 1250 ms-1 is given. Thus
F = ( m / t )v = 275×1250 = 344000 N.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A 2500-kg gondola car
traveling at 3.0 ms-1 has 1500-kg
of sand dropped into it as it travels
by. Find the initial momentum of
the system.
SOLUTION: The system consists of sand and car:
p0,car = mcar v0,car = 2500(3) = 7500 kgms-1.
p0,sand = msandv0,sand = 1500(0) =
0 kgms-1.
p0 = p0,car + p0,sand = 0 + 7500 kgms-1 = 7500 kgms-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A 2500-kg gondola car
traveling at 3.0 ms-1 has 1500-kg
of sand dropped into it as it travels
by. Find the final speed of
the system.
SOLUTION: The initial and final momentums are equal:
p0 = 7500 kgms-1 = pf.
pf = (msand + mcar) vf = (2500 + 1500) vf = 4000 vf.
7500 = 4000 vf  vf = 1.9 ms-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A 2500-kg gondola car
traveling at 3.0 ms-1 has 1500-kg
of sand dropped into it as it travels
by. If the dump lasts 4.5 s, what is
the average force on the car?
SOLUTION: Use Fnet = p / t:
p0 = 7500 kgms-1 = pf.
pf = (msand + mcar) vf = (2500 + 1500) vf = 4000 vf.
7500 = 4000 vf  vf = 1.9 ms-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A 12-kg block of ice is struck by a hammer
so that it breaks into two pieces. The 4.0-kg piece
travels travels at +16 m s-1 in the x-direction. What is the
velocity of the other piece?
8 4
SOLUTION: Make before/after sketches!
The initial momentum of the two is 0.
4 16
8
v
From p = CONST we have p0 = pf.
Since p = mv, we see that
(8 + 4)(0) = 8v + 4(16)  v = -8.0 m s-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
before
25
0
730
1800
after
730
+1800
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A 730-kg Smart Car traveling at 25 m s-1 (xdir) collides with a stationary 1800-kg Dodge Charger.
The two vehicles stick together. Find their velocity
immediately after the collision.
SOLUTION: Make sketches!
p0 = pf so that (730)(25) + 1800(0) = (730 + 1800) vf.
18250 = 2530 vf  vf = 18250 / 2530 = 7.2 m s-1.
vf
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A loaded Glock-22,
having a mass of 975 g, fires
a 9.15-g bullet with a muzzle
velocity of 300 ms-1.
Find the gun’s recoil velocity.
SOLUTION: Use p0 = pf. Then
p0 = pGlock,f + pbullet,f
975(0) = (975 – 9.15)v + (9.15)(-300)
0 = 965.85 v – 2745
v = 2745 / 965.85 = 2.84 m s-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
EXAMPLE: A loaded Glock-22,
having a mass of 975 g, fires
a 9.15-g bullet with a muzzle
velocity of 300 ms-1.
Find the change in kinetic energy
of the gun/bullet system.
SOLUTION: Use EK = (1/2)mv 2 so EK0 = 0 J. Then
EKf = (1/2)(0.975 – 0.00915)2.842 + (1/2)(0.00915)3002
= 416 J.
EK = EKf – EK0 = 416 – 0 = 416 J.
Topic 2: Mechanics
2.4 – Momentum and impulse
Conservation of linear momentum
If Fext = 0 then p = CONST
conservation of
linear momentum
F
EXAMPLE:
How do the ailerons on a plane’s
wing cause it to roll?
SOLUTION:
F
Note that the ailerons oppose each other.
In this picture the right aileron deflects air downward.
Conserving momentum, the right wing dips upward.
In this picture the left aileron deflects air upward.
Conserving momentum, the left wing dips downward.
Topic 2: Mechanics
2.4 – Momentum and impulse
Comparing elastic collisions and inelastic collisions
In an elastic collision, kinetic energy is conserved (it
does not change). Thus EK,f = EK,0.
EXAMPLE:
Two billiard balls colliding in such a way that the speeds
of the balls in the system remain unchanged.
The red ball has the same speed as the white ball…
Both balls have same speeds both before and after…
Topic 2: Mechanics
2.4 – Momentum and impulse
Comparing elastic collisions and inelastic collisions
In an inelastic collision, kinetic energy is not
conserved (it does change). Thus EK,f ≠ EK,0.
EXAMPLE:
A baseball and a hard wall colliding in such a way that
the speed of the ball changes.
Topic 2: Mechanics
2.4 – Momentum and impulse
Comparing elastic collisions and inelastic collisions
In a completely inelastic collision the colliding
bodies stick together and end up with the same
velocities, but different from the originals. EK,f ≠ EK,0.
EXAMPLE:
Two objects colliding and sticking together.
u1
v
u2
v
The train cars hitch and move as one body…
The cars collide and move (at first) as one body…
Topic 2: Mechanics
2.4 – Momentum and impulse
Comparing elastic collisions and inelastic collisions
 An explosion is similar to a completely inelastic
collision in that the bodies were originally stuck together
and began with the same velocities. EK,f ≠ EK,0.
EXAMPLE:
Objects at rest suddenly separating into two pieces.
A block of ice broken in two by a hammer stroke…
A bullet leaving a gun
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
conservation of
If Fext = 0 then p = CONST
linear momentum
EXAMPLE: Two train cars having equal masses of 750
kg and velocities u1 = 10. m s-1 and u2 = 5.0 m s-1 collide
and hitch together. What is their final speed?
u1
v
u2
v
SOLUTION: Use momentum conservation p0 = pf. Then
p1,0 + p2,0 = p1,f + p2,f
mu1 + mu2 = mv + mv
m(u1 + u2) = 2mv
10 + 5 = 2v  v = 7.5 m s-1.
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
conservation of
If Fext = 0 then p = CONST
linear momentum
EXAMPLE: Two train cars having equal masses of 750
kg and velocities u1 = 10. m s-1 and u2 = 5.0 m s-1 collide
and hitch together. Find the change in kinetic energy.
u1
v
u2
v
SOLUTION: Use EK = (1/2) mv 2. Then
EK,f = (1/2) (m + m) v 2
= (1/2) (750 + 750) 7.5 2 = 42187.5 J.
EK,0 = (1/2) (750) 10 2 + (1/2) (750) 5 2 = 46875 J.
EK = EK,f – EK,0 = 42187.5 – 46875 = - 4700 J.
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
conservation of
If Fext = 0 then p = CONST
linear momentum
EXAMPLE: Two train cars having equal masses of 750
kg and velocities u1 = 10. m s-1 and u2 = 5.0 m s-1 collide
and hitch together. Determine the type of collision.
u1
v
u2
v
SOLUTION:
Since EK,f ≠ EK,0, this is an inelastic collision.
Since the two objects travel as one (they are stuck
together) this is also a completely inelastic collision.
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
conservation of
If Fext = 0 then p = CONST
linear momentum
EXAMPLE: Two train cars having equal masses of 750
kg and velocities u1 = 10. m s-1 and u2 = 5.0 m s-1 collide
and hitch together. Was mechanical energy conserved?
u1
v
u2
v
SOLUTION:
Mechanical energy E = EK + EP.
Since the potential energy remained constant and the
kinetic energy decreased, the mechanical energy was
not conserved.
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
conservation of
If Fext = 0 then p = CONST
linear momentum
EXAMPLE: Two train cars having equal masses of 750
kg and velocities u1 = 10. m s-1 and u2 = 5.0 m s-1 collide
and hitch together. Was total energy conserved?
u1
v
u2
v
SOLUTION:
Total energy is always conserved.
The loss in mechanical energy is EK = - 4700 J.
The energy lost is mostly converted to heat (there is
some sound, and possibly light, but very little).
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block of wood
resting on a wood floor. How fast is the block/bullet
combo moving immediately after collision?
SOLUTION:
If we consider the bullet-block combo as our system,
there are no external forces in the x-direction at
collision. Thus pf = p0 so that
the bullet and the block
mvf + MVf = mvi
+ MVi
move at the same
.02v + 4 v = (.02)(300) + 4(0)
speed after collision
4.02v = 6
(completely inelastic)
v = 1.5 m/s
f
Topic 2: Mechanics
2.4 – Momentum and impulse
s
Quantitatively analysing inelastic collisions
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block of wood
resting on a wood floor. The block/bullet combo slides 6
m before coming to a stop. Find the friction f between
the block and the floor.
SOLUTION: Use the work-kinetic energy theorem:
∆EK = W
(1/2)mv 2 – (1/2)mu 2 = f s cos 
(1/2)(4.02)(0)2 – (1/2)(4.02)(1.5)2 = f (6) cos 180°
- 4.5225 = - 6f
f = - 4.5225 / - 6
f = 0.75 N.
f
Topic 2: Mechanics
2.4 – Momentum and impulse
s
Quantitatively analysing inelastic collisions
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block of wood
resting on a wood floor. The block/bullet combo slides 6
m before coming to a stop. Find the dynamic friction
coefficient µd between the block and the floor.
R
SOLUTION: Use f = µdR:
Make a free-body diagram to
f
find R:
W
Note that R = W = mg
= (4.00 + 0.020)(10) = 40.2 N.
Thus
µ = f / R = 0.75 / 40.2 = 0.19.
F s
Topic 2: Mechanics
2.4 – Momentum and impulse
Quantitatively analysing inelastic collisions
EXAMPLE: Suppose a .020-kg bullet traveling
horizontally at 300. m/s strikes a 4.0-kg block of wood
resting on a wood floor. If the bullet penetrates .060 m of
the block, find the average force F acting on it during its
collision.
SOLUTION: Use the work-kinetic energy theorem on
only the bullet:
∆EK = W
(1/2)mv 2 – (1/2)mu 2 = F s cos 
(1/2)(.02)(1.5)2 – (1/2)(.02)(300)2 = - F (.06)
- 900 = - 0.06F
F = 15000 n.