x2 t06 02 resisted motion (2013)
TRANSCRIPT
Resisted Motion
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
mg
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
Rmg
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
R
Rmgxm
mg
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
R
Rmgxm
mRgx
mg
Resisted MotionResistance is ALWAYS in the OPPOSITE direction to the motion.
(Newton’s 3rd Law)Case 1 (horizontal line)
xm
R
Rxm
mRx
Case 2 (upwards motion)
xm
R
Rmgxm
mRgx
mg NOTE: greatest height still occurs when v = 0
Case 3 (downwards motion)
Case 3 (downwards motion)
Case 3 (downwards motion)
xm
Case 3 (downwards motion)
xm mg
Case 3 (downwards motion)
xm
R
mg
Case 3 (downwards motion)
xm
RRmgxm
mg
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
xm
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
xm
mg
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
xm
mg 2kv
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
xm
mg 2kv
uvx ,0
Case 3 (downwards motion)
xm
RRmgxm
mRgx
mg NOTE:0when
occursvelocity terminalx
e.g. (i) A particle is projected vertically upwards with a velocity of u m/s in a resisting medium.Assuming that the retardation due to this resistance is equal tofind expressions for the greatest height reached and the time taken to reach that height.
2kv
xm
mg 2kv 2
2
vmkgx
kvmgxm
uvx ,0
mkvmg
dxdvv
2
mkvmg
dxdvv
2
2kvmgmv
dvdx
mkvmg
dxdvv
2
2kvmgmv
dvdx
u
o
u
kvmgkvdv
km
kvmgvdvmx
02
2
22
mkvmg
dxdvv
2
2kvmgmv
dvdx
u
o
u
kvmgkvdv
km
kvmgvdvmx
02
2
22
ukvmgk
m0
2log2
mkvmg
dxdvv
2
2kvmgmv
dvdx
u
o
u
kvmgkvdv
km
kvmgvdvmx
02
2
22
ukvmgk
m0
2log2
mgku
km
mgkumg
km
mgkumgk
m
2
2
2
1log2
log2
loglog2
mkvmg
dxdvv
2
2kvmgmv
dvdx
u
o
u
kvmgkvdv
km
kvmgvdvmx
02
2
22
ukvmgk
m0
2log2
mgku
km
mgkumg
km
mgkumgk
m
2
2
2
1log2
log2
loglog2
metres 1log2
is
height greatest the2
mgku
km
2vmkgx
2vmkgx
mkvmg
dtdv 2
2vmkgx
mkvmg
dtdv 2
uu
vk
mgdv
km
kvmgdvmt
0 2
0
2
2vmkgx
mkvmg
dtdv 2
uu
vk
mgdv
km
kvmgdvmt
0 2
0
2
u
vmgk
mgk
km
0
1tan
2vmkgx
mkvmg
dtdv 2
uu
vk
mgdv
km
kvmgdvmt
0 2
0
2
u
vmgk
mgk
km
0
1tan
umgk
kgm
umgk
kgm
1
11
tan
0tantan
2vmkgx
mkvmg
dtdv 2
uu
vk
mgdv
km
kvmgdvmt
0 2
0
2
u
vmgk
mgk
km
0
1tan
umgk
kgm
umgk
kgm
1
11
tan
0tantan
height.greatest reach the to
seconds tan it takes 1
u
mgk
kgm
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
v81
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
v81 vgx
vgx
4018155
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
v81 vgx
vgx
4018155
4040 vg
dtdv
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
v81 vgx
vgx
4018155
4040 vg
dtdv
v
vgdvt
0 4040
(ii) A body of mass 5kg is dropped from a height at which the gravitational acceleration is g.Assuming that air resistance is proportional to speed v, the constant of proportion being , find;
81
a) the velocity after time t.
x5 g5
v81 vgx
vgx
4018155
4040 vg
dtdv
v
vgdvt
0 4040
vgg
gvgvg v
4040log40
40log40log4040log40 0
vg
gt40
40log40
vg
gt40
40log40
40
4040 t
evg
g
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity0when occurs velocity terminal x
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity0when occurs velocity terminal x
gv
vg
404010i.e.
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity0when occurs velocity terminal x
gv
vg
404010i.e.
OR
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity0when occurs velocity terminal x
gv
vg
404010i.e.
OR
g
egvt
tt
40
140limlim 40
vg
gt40
40log40
40
4040 t
evg
g
40
4040 t
eg
vg
40
40
40
140
4040
4040
t
t
t
egv
gegv
gevg
b) the terminal velocity0when occurs velocity terminal x
gv
vg
404010i.e.
OR
g
egvt
tt
40
140limlim 40
m/s40isvelocity terminal g
c) The distance it has fallen after time t
c) The distance it has fallen after time t
40140t
egdtdx
c) The distance it has fallen after time t
40140t
egdtdx
t t
dtegx0
40140
c) The distance it has fallen after time t
40140t
egdtdx
t t
dtegx0
40140
tt
etgx0
404040
c) The distance it has fallen after time t
40140t
egdtdx
t t
dtegx0
40140
tt
etgx0
404040
ggegtx
etgx
t
t
1600160040
4004040
40
40
c) The distance it has fallen after time t
40140t
egdtdx
t t
dtegx0
40140
tt
etgx0
404040
ggegtx
etgx
t
t
1600160040
4004040
40
40
Exercise 8C; 2, 4, 5, 6, 8, 10, 13, 16