physics
Content
Review:
Lecture 6 More Problems and 3D kinematics
For motion under gravity (constant acceleration) we found:
a = const v = v0 + a t r = r0 + v0 t + 1/2 a t2
t
v y = v0 y − gt y = y0 + v 0 y t −
Alice v0
v0
y
ay = −g
Bill
v
1 2 gt 2
t
v2 − v02y = −2g( y − y0 )
a t
y=0
x Physics 1301: Lecture 6, Pg 1
Physics 1301: Lecture 6, Pg 2
ICQ 1D free fall
ICQ 1D free fall
What is the speed of Bill’s ball as it passes back through height H? Since the motion up and back down is symmetric, intuition should tell you that v = -v0 We can prove that your intuition is correct:
Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively. Ignoring air resistance, which of the following is true:
2
(a)
vA < vB
(b) vA = vB
Alice v0
v0
v 2 − v0 = 2a( y − y0 )
(c) vA > vB
v 2 − v02 = 2(−g)(H − H ) = 0
Bill
v0 H
vA
Physics 1301: Lecture 6, Pg 4
Remember: How to solve problems
ICQ 1D free fall
Read and understand the problem Give a "Physics Description" of the event, draw diagram Plan the Solution, write down the equations Solve the equations and plug in numbers Check and Evaluate
We can also just use the equation directly: 2
v 2 − v0 = 2a( y − y0 ) Alice:
v 2 − v02 = 2(−g )(0 − H )
Bill:
v 2 − v 02 = 2( −g )(0 − H )
Bill
This looks just like Bill threw the ball down with speed v0, so the speed at the bottom should be the same as Alice’ Alice’s ball. y=0
Physics 1301: Lecture 6, Pg 3
v0
v = -v0 H
vB
UIUC
Alice v0
v = ± v02
Bill
Watch your units ! ¾Always check the units of your answer, and carry the units along with your numbers during the calculation. Understand the limits ! ¾Many equations we use are special cases of more general laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).
same !!
Of course it takes longer for Bill’s ball to reach the ground but its velocity is the same y=0 Physics 1301: Lecture 6, Pg 5
Physics 1301: Lecture 6, Pg 6
Page 1
The Sprint missile
The Sprint missile
The Sprint missile, designed to destroy incoming ballistic missiles, can accelerate at 100 g. If an ICBM is detected at an altitude of 100 km moving straight down at a constant speed of 3 x 104 km/h and the Sprint missile is launched to intercept it, at what time and altitude will the interception take place? (Note: Neglect the acceleration due to gravity. ) What’s happening? – 1D motion with constant acceln Set up the physical picture
x=1/2at2+vot+x0
1 2
They meet when
as=100g
XS=0
x
Vs=0
x S = x ICBM
1 2 a s t = H − v ICBM t 2 1 2 a s t + v ICBM t − H = 0 2
t=0
Physics to use
1 2 1 a s t + v 0 s t + x0 s = a s t 2 2 2 v0s=0,x0s=0
The ICBM x ICBM = H - v ICBM t + a ICBM t 2
XICBM= H = 100km
4 They will intercept vICBM = -3x10 km/h a when xs=xICBM ICBM=0
Position with constant acceleration
The missile x S =
Physics 1301: Lecture 6, Pg 7
The Sprint missile 1 a s t 2 + v ICBM t − H = 0 2
What to do if you have problems with the problems
Remember
Firstly look at the text and/or the lectures for a similar problem and try to do it the same way Secondly look at the solutions. They are on the class web. They show some of the steps in getting to the solutions, it will probably help to follow them If you still have a problem bring your workings to me or your TA and we will see where you are going wrong
a 2 − v ICBM ± v ICBM − 4 × s × − H 2 t= as
a s = 100g = 981m/s H = 100km = 105 m
Height
x Physics 1301: Lecture 6, Pg 8
−b ± b 2 − 4ac t= 2a
Now put in the v ICBM = 3 × 10 4 km/h = 8.33 × 103 m/s numbers
aICBM=0
UNITS!
2
t=8.12 s
x i = H-v ICBM t = 32 .4 km
Is it reasonable? Significant figures
x
Physics 1301: Lecture 6, Pg 9
Physics 1301: Lecture 6, Pg 10
ICQ: Acceleration
ICQ: Acceleration
An object falling near the surface of the earth has a constant acceleration due to gravity of approximately 9.8 m/s2. This means that:
An object falling near the surface of the earth has a constant acceleration due to gravity of approximately 9.8 m/s2. This means that:
a) The object falls 9.8m during the first second of its motion b) The object falls 9.8m during each second of its motion c) The speed of the object increases by 9.8m/s during each second of its motion d) The acceleration of the object increases by 9.8m/s2 during each second of its motion
a) The object falls 9.8m during the first second of its motion b) The object falls 9.8m during each second of its motion c) The speed of the object increases by 9.8m/s during each second of its motion d) The acceleration of the object increases by 9.8m/s2 during each second of its motion
v=v0+at Physics 1301: Lecture 6, Pg 11
x=x0+v0t+1/2at2 Physics 1301: Lecture 6, Pg 12
Page 2
The two ball problem
The two ball problem Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur? Initial Conditions, define origin at the ground and x upwards: xA0 = H xB0 = 0 vA0 = 0 vB0 = vB0 H aA =aB= -g
··· At what fraction of the height of the building does the collision occur? Conditions when they collide at time t: xA = xB vA = -2vB
A
A position
xA = H - (½)gt 2 xB = vB0t - (½)gt 2
Physics: linear motion, constant acceleration Conditions at collision: xA = xB vA = -2vB
Apply condition on x xA = xB
B
⇒
B
H = vB0t
time
Physics 1301: Lecture 6, Pg 13
Physics 1301: Lecture 6, Pg 14
The two ball problem
3-D Kinematics
xA = H - (½)gt 2 xB = vB0t - (½)gt 2 H = vB0t
Expressions for vA and vB vA = -gt vB = vB0 - gt
Ball Dropping
Equations of motion:
The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as vectors:
v
Set -vA = 2vB and solve for t gt = 2 (vB0 - gt)
vB
t = 2vB0/3g
vA
Then, eliminate vB0 t = 2(H/t)/3g = 2H/(3gt) ⇒ (½)gt From our calculation of xA xA = H - (½)gt 2 = H-H/3 = 2H/3
2
r = xi+yj+zk v = vx i + vy j + v z k
(ii , j , k unit vectors)
a = ax i + ay j + az k
t
We have already seen the 1-D kinematics equations:
= H/3
x = x(t )
Units? Is it reasonable?
v=
dx dt
a=
dv d 2 x = dt dt 2
They meet at 2/3 of the height of the building Physics 1301: Lecture 6, Pg 15
Physics 1301: Lecture 6, Pg 16
3-D Kinematics
3-D Kinematics So for constant acceleration we can integrate to get:
In 3-D, the three components are independent, we simply apply the 1-D equations to each of the component equations.
x = x(t )
vx =
ax =
dx dt
vy =
d2x dt
y = y( t )
2
ay =
dy dt
vz =
d2y dt
2
a = const v = v0 + a t r = r0 + v0 t + 1/2 a t2
z = z( t )
az =
dz dt
(where a, v, v0, r, r0, are all vectors)
d 2z dt
2
Each component (x,y,z) also follows the same equations
Which can be combined into the vector equations: r = r(t)
v = dr (t)/ dt
a=
ax=const vx=v0x+axt x=x0+v0x+ ½ axt2
d2r(t)/dt2 r
ay=const vy=v0y+ayt y=y0+v0y+ ½ ayt2
az=const vz=v0z+azt z=z0+v0z+azt2
UIUC Physics 1301: Lecture 6, Pg 17
Physics 1301: Lecture 6, Pg 18
Page 3
2-D Kinematics
Simultaneous fall Two balls are held at height h (2m) from the ground. ¾Ball 1 is released from rest. ¾Ball 2 is simultaneously ejected with velocity v2 to the right Which ball hits the ground first?
Most 3-D problems can be reduced to 2-D problems when the acceleration is constant: ¾Choose y axis to be along direction of acceleration ¾Choose x axis to be along the “other” direction of motion
1 a x t 2 = x10 2 1 1 2 y1 = y10 + v1 y 0 t + a y t = h − gt 2 2 2 1 x 2 = x 20 + v 2 x 0 t + a x t 2 = x 20 + v 2 t 2 1 y 2 = y 20 + v 2 y 0 t + a y t 2 = h − 1 gt 2 2 2 x 1 = x 10 + v 1 x 0 t +
Example: Example Throwing a ball (neglecting wind, air resistance) ¾Acceleration is constant (gravity) ¾Choose y axis up: ay = -g ¾Choose x axis along the ground in the direction of the throw
y1 = y2 , therefore they hit simultaneously
y
h
x
It doesn’t matter what h or v2 are! Physics 1301: Lecture 6, Pg 19
Physics 1301: Lecture 6, Pg 20
Components of motion
Homework Read Chapter 3 of Fishbane Do problems, Fishbane chapter 2 #5,9,19,3349, 62, 76 if you haven’t already done them.
x and y (and z) components of the motion are INDEPENDENT ¾The two balls fall the same distance in y under the same acceleration (gravity) ¾They take the same time to hit the floor It doesn’t matter what happens in the x direction (providing it doesn’t hit anything before it hits the floor and we can ignore the curvature of the earth) In any problem you can (and should) always resolve the motion along orthogonal axes (axes at 900) and write down independent equations in the x,y,z directions You can solve the equations independently Provides extra equations to solve for unknowns
Physics 1301: Lecture 6, Pg 21
Physics 1301: Lecture 6, Pg 22
Page 4
Lecture 6 More Problems and 3D kinematics
For motion under gravity (constant acceleration) we found:
a = const v = v0 + a t r = r0 + v0 t + 1/2 a t2
t
v y = v0 y − gt y = y0 + v 0 y t −
Alice v0
v0
y
ay = −g
Bill
v
1 2 gt 2
t
v2 − v02y = −2g( y − y0 )
a t
y=0
x Physics 1301: Lecture 6, Pg 1
Physics 1301: Lecture 6, Pg 2
ICQ 1D free fall
ICQ 1D free fall
What is the speed of Bill’s ball as it passes back through height H? Since the motion up and back down is symmetric, intuition should tell you that v = -v0 We can prove that your intuition is correct:
Alice and Bill are standing at the top of a cliff of height H. Both throw a ball with initial speed v0, Alice straight down and Bill straight up. The speed of the balls when they hit the ground are vA and vB respectively. Ignoring air resistance, which of the following is true:
2
(a)
vA < vB
(b) vA = vB
Alice v0
v0
v 2 − v0 = 2a( y − y0 )
(c) vA > vB
v 2 − v02 = 2(−g)(H − H ) = 0
Bill
v0 H
vA
Physics 1301: Lecture 6, Pg 4
Remember: How to solve problems
ICQ 1D free fall
Read and understand the problem Give a "Physics Description" of the event, draw diagram Plan the Solution, write down the equations Solve the equations and plug in numbers Check and Evaluate
We can also just use the equation directly: 2
v 2 − v0 = 2a( y − y0 ) Alice:
v 2 − v02 = 2(−g )(0 − H )
Bill:
v 2 − v 02 = 2( −g )(0 − H )
Bill
This looks just like Bill threw the ball down with speed v0, so the speed at the bottom should be the same as Alice’ Alice’s ball. y=0
Physics 1301: Lecture 6, Pg 3
v0
v = -v0 H
vB
UIUC
Alice v0
v = ± v02
Bill
Watch your units ! ¾Always check the units of your answer, and carry the units along with your numbers during the calculation. Understand the limits ! ¾Many equations we use are special cases of more general laws. Understanding how they are derived will help you recognize their limitations (for example, constant acceleration).
same !!
Of course it takes longer for Bill’s ball to reach the ground but its velocity is the same y=0 Physics 1301: Lecture 6, Pg 5
Physics 1301: Lecture 6, Pg 6
Page 1
The Sprint missile
The Sprint missile
The Sprint missile, designed to destroy incoming ballistic missiles, can accelerate at 100 g. If an ICBM is detected at an altitude of 100 km moving straight down at a constant speed of 3 x 104 km/h and the Sprint missile is launched to intercept it, at what time and altitude will the interception take place? (Note: Neglect the acceleration due to gravity. ) What’s happening? – 1D motion with constant acceln Set up the physical picture
x=1/2at2+vot+x0
1 2
They meet when
as=100g
XS=0
x
Vs=0
x S = x ICBM
1 2 a s t = H − v ICBM t 2 1 2 a s t + v ICBM t − H = 0 2
t=0
Physics to use
1 2 1 a s t + v 0 s t + x0 s = a s t 2 2 2 v0s=0,x0s=0
The ICBM x ICBM = H - v ICBM t + a ICBM t 2
XICBM= H = 100km
4 They will intercept vICBM = -3x10 km/h a when xs=xICBM ICBM=0
Position with constant acceleration
The missile x S =
Physics 1301: Lecture 6, Pg 7
The Sprint missile 1 a s t 2 + v ICBM t − H = 0 2
What to do if you have problems with the problems
Remember
Firstly look at the text and/or the lectures for a similar problem and try to do it the same way Secondly look at the solutions. They are on the class web. They show some of the steps in getting to the solutions, it will probably help to follow them If you still have a problem bring your workings to me or your TA and we will see where you are going wrong
a 2 − v ICBM ± v ICBM − 4 × s × − H 2 t= as
a s = 100g = 981m/s H = 100km = 105 m
Height
x Physics 1301: Lecture 6, Pg 8
−b ± b 2 − 4ac t= 2a
Now put in the v ICBM = 3 × 10 4 km/h = 8.33 × 103 m/s numbers
aICBM=0
UNITS!
2
t=8.12 s
x i = H-v ICBM t = 32 .4 km
Is it reasonable? Significant figures
x
Physics 1301: Lecture 6, Pg 9
Physics 1301: Lecture 6, Pg 10
ICQ: Acceleration
ICQ: Acceleration
An object falling near the surface of the earth has a constant acceleration due to gravity of approximately 9.8 m/s2. This means that:
An object falling near the surface of the earth has a constant acceleration due to gravity of approximately 9.8 m/s2. This means that:
a) The object falls 9.8m during the first second of its motion b) The object falls 9.8m during each second of its motion c) The speed of the object increases by 9.8m/s during each second of its motion d) The acceleration of the object increases by 9.8m/s2 during each second of its motion
a) The object falls 9.8m during the first second of its motion b) The object falls 9.8m during each second of its motion c) The speed of the object increases by 9.8m/s during each second of its motion d) The acceleration of the object increases by 9.8m/s2 during each second of its motion
v=v0+at Physics 1301: Lecture 6, Pg 11
x=x0+v0t+1/2at2 Physics 1301: Lecture 6, Pg 12
Page 2
The two ball problem
The two ball problem Ball A is dropped from the top of a building at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what fraction of the height of the building does the collision occur? Initial Conditions, define origin at the ground and x upwards: xA0 = H xB0 = 0 vA0 = 0 vB0 = vB0 H aA =aB= -g
··· At what fraction of the height of the building does the collision occur? Conditions when they collide at time t: xA = xB vA = -2vB
A
A position
xA = H - (½)gt 2 xB = vB0t - (½)gt 2
Physics: linear motion, constant acceleration Conditions at collision: xA = xB vA = -2vB
Apply condition on x xA = xB
B
⇒
B
H = vB0t
time
Physics 1301: Lecture 6, Pg 13
Physics 1301: Lecture 6, Pg 14
The two ball problem
3-D Kinematics
xA = H - (½)gt 2 xB = vB0t - (½)gt 2 H = vB0t
Expressions for vA and vB vA = -gt vB = vB0 - gt
Ball Dropping
Equations of motion:
The position, velocity, and acceleration of a particle in 3 dimensions can be expressed as vectors:
v
Set -vA = 2vB and solve for t gt = 2 (vB0 - gt)
vB
t = 2vB0/3g
vA
Then, eliminate vB0 t = 2(H/t)/3g = 2H/(3gt) ⇒ (½)gt From our calculation of xA xA = H - (½)gt 2 = H-H/3 = 2H/3
2
r = xi+yj+zk v = vx i + vy j + v z k
(ii , j , k unit vectors)
a = ax i + ay j + az k
t
We have already seen the 1-D kinematics equations:
= H/3
x = x(t )
Units? Is it reasonable?
v=
dx dt
a=
dv d 2 x = dt dt 2
They meet at 2/3 of the height of the building Physics 1301: Lecture 6, Pg 15
Physics 1301: Lecture 6, Pg 16
3-D Kinematics
3-D Kinematics So for constant acceleration we can integrate to get:
In 3-D, the three components are independent, we simply apply the 1-D equations to each of the component equations.
x = x(t )
vx =
ax =
dx dt
vy =
d2x dt
y = y( t )
2
ay =
dy dt
vz =
d2y dt
2
a = const v = v0 + a t r = r0 + v0 t + 1/2 a t2
z = z( t )
az =
dz dt
(where a, v, v0, r, r0, are all vectors)
d 2z dt
2
Each component (x,y,z) also follows the same equations
Which can be combined into the vector equations: r = r(t)
v = dr (t)/ dt
a=
ax=const vx=v0x+axt x=x0+v0x+ ½ axt2
d2r(t)/dt2 r
ay=const vy=v0y+ayt y=y0+v0y+ ½ ayt2
az=const vz=v0z+azt z=z0+v0z+azt2
UIUC Physics 1301: Lecture 6, Pg 17
Physics 1301: Lecture 6, Pg 18
Page 3
2-D Kinematics
Simultaneous fall Two balls are held at height h (2m) from the ground. ¾Ball 1 is released from rest. ¾Ball 2 is simultaneously ejected with velocity v2 to the right Which ball hits the ground first?
Most 3-D problems can be reduced to 2-D problems when the acceleration is constant: ¾Choose y axis to be along direction of acceleration ¾Choose x axis to be along the “other” direction of motion
1 a x t 2 = x10 2 1 1 2 y1 = y10 + v1 y 0 t + a y t = h − gt 2 2 2 1 x 2 = x 20 + v 2 x 0 t + a x t 2 = x 20 + v 2 t 2 1 y 2 = y 20 + v 2 y 0 t + a y t 2 = h − 1 gt 2 2 2 x 1 = x 10 + v 1 x 0 t +
Example: Example Throwing a ball (neglecting wind, air resistance) ¾Acceleration is constant (gravity) ¾Choose y axis up: ay = -g ¾Choose x axis along the ground in the direction of the throw
y1 = y2 , therefore they hit simultaneously
y
h
x
It doesn’t matter what h or v2 are! Physics 1301: Lecture 6, Pg 19
Physics 1301: Lecture 6, Pg 20
Components of motion
Homework Read Chapter 3 of Fishbane Do problems, Fishbane chapter 2 #5,9,19,3349, 62, 76 if you haven’t already done them.
x and y (and z) components of the motion are INDEPENDENT ¾The two balls fall the same distance in y under the same acceleration (gravity) ¾They take the same time to hit the floor It doesn’t matter what happens in the x direction (providing it doesn’t hit anything before it hits the floor and we can ignore the curvature of the earth) In any problem you can (and should) always resolve the motion along orthogonal axes (axes at 900) and write down independent equations in the x,y,z directions You can solve the equations independently Provides extra equations to solve for unknowns
Physics 1301: Lecture 6, Pg 21
Physics 1301: Lecture 6, Pg 22
Page 4
