Download 7.1 The Impulse-Momentum Theorem

7.1 The Impulse-Momentum Theorem
TRSP Fig. 7-1b: Force on a baseball.
Definition of Impulse: the impulse of a force is the
product of the average force F and the time interval
∆t during which the force acts:
Impulse = F ∆t
(7.1)
Impulse is a vector quantity and has the same direction
as the average force.
SI Unit of Impulse: newton·second ( N · s)
Definition of Linear Momentum: the linear momen→
tum −
p of an object is the product of the object’s
→
v:
mass m and velocity −
−
→
→
p = m−
v
(7.2)
TRSP Fig. 7.4: average force and velocity change
a=
→
−
→−−
v
f v0
∆t
From N2: F
Ã−
!
→
−
→
→−−
−v
→
m
vf −v0
m−
v
0
f
=m
=
∆t
∆t
(7.3)
Impulse-Momentum Theorem: When a net
force acts on an object, the impulse of the net
force is equal to the change in momentum of
the object:
−
→
F ∆t = m−
v→
f − mv0
(7.4)
Impulse = Change in momentum
Ex. 1: Hitting a baseball (mass, m = 0.14 kg),
−
→ = 58 m/ s, ∆t = 1.6 ×
−
→ = −38 m/ s, v
initially v
0
f
10−3 s
−
→ − m−
→
(a) impulse = mv
v
0
f
= 0.14 kg [58 m/ s − (−38 m/ s)] = . 14 ( kg) 96.0 m
s =
13. 44 ( kg) m
s
(b) F =
N
impulse
∆t
=
13. 44( kg) m
s
1.6×10−3 s
= 8400.0 ( kg) sm2 or
Ex.2: Rain Storm:
−
→ = −15 m/ s, rate of rain is 0.060 kg/ s. Find averv
0
age force on the car if raindrops come to rest.
average force on the rain
F=
→−m
−−v
→
m−
v
0
f
∆t
=−
³
m
∆t
´
−
→
v
0
= 0.060 kg/ s × −15 m/ s = −. 9 ( kg) m
s or N
Force on roof = +. 9 ( kg) m
s or N
CEx. 3: Hailstones versus Raindrops: Hailstones rebound giving a greater force.
Problems 8, 10.
7.2 The Principle of Conservation of
Linear Momentum
TRSP Fig. 7.7 Concepts at a Glance
On page 195/6 Cutnell and Johnson prove that for an
isolated system (no external force, e.g. gravity):
→
−
→ −
Pf = Pi
(7.7)
−
→
−→
where Pf = m1−
v→
f1 + m2 vf 2,
−
→
−
→
and Pi = m1−
v→
i1 + m2 vi2
Principle of Conservation of Linear Momentum: The
total linear momentum of an isolated system remains
constant (is conserved). An isolated system is one for
which the vector sum of the external forces acting on
the system is zero.
CEx. 4: READ!
Ex. 5: Assembling a Freight Train:
m1 = 65 × 103 kg, v01 = +0.80 m/ s is overtaken by
m2 = 92 × 103 kg, v01 = +1.20 m/ s
Find the common velocity when they are coupled:
(m1 + m2) vf = m1v01 + m2v02
2 v02 → 1.0 m/ s
which solves for vf = m1vm01+m
1+ m2
Ex.6: Ice Skaters: TRSP Fig. 7.11: simply a case of
total momentum being zero - READ
[CJ4: CEx. 5. Firing Blanks]
Problems 20, 46
REASONING STRATEGY: page 200 - READ
7.3 Collisions in One Dimension
Elastic collisions
Inelastic collisions
TRSP Fig. 7.13
Ex. 7: 1D collision: m1 = 0.250 kg, v01 = 5.00 m/ s
collides head-on with m2 = 0.800 kg, v02 = 0 (at
rest):
Fig. 7.12 (DRAW)
m1vf1 + m2vf 2 = m1v01 + 0
1 m v2
2 1 f1
(A)
2 = 1 m v2 + 0
+ 12 m2vf2
2 1 01
(B)
Generalization of 1D to include an initially moving m2:
m1vf1 + m2vf 2 = m1v01 + m2v02
1 m v2
2 1 f1
(A’)
2 = 1 m v2 + 1 m v2
+ 12 m2vf2
2 1 01
2 2 02
(B’)
SIMPLIFY THE VARIABLES:
m1 = a, m2 = b, vf 1 = x, vf2 = y, v01 = c, v02 =
d:
Solve A’: ax + by = ac + bd,
Solution is :
n
y=
− xa−ac−bd
b
o
,
eliminate y in B’ after dropping the ”halves”:
ax2 + by 2 = ac2 + bd2 gives a quadratic in x:
ax2
+b
³
´2
xa−ac−bd
−
b
= ac2 + bd2,
Solutions are : {x = c} ,
n
and x =
2bd+ac−cb
b+a
o
The first solution is incompatible with a moving m2
after the collision, hence we choose the second solution to substitute into the momentum equation, A’:
ax + by = ac → a
Solution is :
n
y=
³
2bd+ac−cb
b+a
2a −d+c
b+a
o
´
+ by = ac,
,
NOW set d = v02 = 0:
Thus vf 1 =
³
m1−m2
m1+m2
´
v01
(7.8a)
and vf 2 =
³
2m1
m1 +m2
´
v01
(7.8b)
Using the numerical values: vf 1 = −2.62 m/ s (so ball
1 rebounds left) and vf 2 = +2.38 m/ s
N.B. if m1 < m2 then vf1 will be negative.
ALSO if m1 = m2 then vf1 will be zero (stationary).
Ex. 8: Ballistic Pendulum:
Fig. 7.15 (DRAW)
bullet: m1 = 0.01 kg , speed v01; block of wood:
m2 = 2.5 kg rises to h = 0.65 m: Find v01.
conserve momentum: (m1 + m2) vf = m1v01(a),
Solution is :
n
v01 =
vf o
(m1 + m2) m
1
conserve energy: (m1 + m2) ghf = 12 (m1 + m2) vf2 (b),
CANCEL common factor, so that:
vf2 = 2ghf
Numerically: vf = 3.57 m/ s and v01 = 896 m/ s.
7.4 Collisions in Two Dimensions
TRSP Fig. 7.16
Ex. 9: 2D collision: DATA IN FIG. 7.15:
x component: Pf x = P0x
³
(7.9a)
´
(15 kg) vf1x + (26 kg) (0.7 m/ s) (cos 35 ◦) =
(15 kg) (0.9 m/ s) (sin 50 ◦) + (26 kg) (0.54 m/ s)
vf 1x = (−14. 909 + 24. 382) /15.0 = . 631 53 m/ s
y component: Pf y = P0x
³
(7.9b)
´
(15 kg) vf1y + (26 kg) (−0.7 m/ s) (sin 35 ◦) =
(15 kg) (−0.9 m/ s) (cos 50 ◦) + 0
vf 1y = (−10. 439 + 8. 677 6) /15.0 = . 117 43 m/ s
The final speed of m1 is:
vf 1 =
³
. 631 532
´1
2
+ . 117 43 2
= . 642 35 m/ s
and its direction has:
θ = arctan
³
. 117 43
. 631 53
´
= . 183 85 rad or 10.5 ◦
7.7 Centre of Mass
(abbreviate: CM)
2 x2 +...
xCM = m1mx1+m
1 +m2 +...
(7.10)
For 2 masses, if m1 = m2 = m ,
then xCM = 12 (x1 + x2).
Symmetric objects have the CM at their geometric
centres.
Velocity of the CM:
1 +m2∆x2
∆xCM = m1∆x
m1 +m2
divide by ∆t:
2 v2
vCM = m1mv1+m
1 +m2
(7.11)