Quiz 1
This is a photograph of an apple and feather free fall in a evacuated chamber. The
apple and feather are released from the top. Suppose the camera opened
flash every 1 s.
(1) Suppose A1 is the distance from t = 0s to t = 1 s, A2 is the distance from t = 1s
to t =2s, A3 is the distance from t = 2s to t = 3s ……
what is the ratio of A1: A2: A3…….
(2) If camera opened flash every 0.06s, is the ratio of the distance between every
0.06 s the same as above or change? Why?
Answer: The ratio is 1:3:5 :7……(2n -1), here n = 1, 2, 3,
• Show it in general case, suppose the time interval is Δt
The distance from 0 –one Δt, A1 = ½*g*(Δt)2
(notice here I consider the distance, so I neglect (-) sign)
The distance from Δt---2 Δt, A2 = ½*g*(2Δt)2 -½*g*(Δt)2
= ½*g*(Δt)2 (22 -1)
Using the same way the distance from 2Δt---3 Δt,
A3 = ½*g*(Δt)2 (32 - 22)
A4 = ½*g*(Δt)2 (42 - 32) …….
An = ½*g*(Δt)2 (n2 – (n-1)2)
= ½*g*(Δt)2 (n2 – n2 +2n-1)
= ½*g*(Δt)2 (2n-1)
Therefore the ratio should be 1:3:5…….(2n-1), the ratio is
independent of Δt
Quiz 2
In a projectile motion, if we measure the maximum
height h and horizontal range R
(1) How to get launch angle θ in terms of h of and R?
(2) A hobby rocket reaches a height of 72.3 m and lands
111 m from the launch point.
What is the angle of launch?
What is the launch speed?
( you need show your work, no work, no points)
We know The maximum height
and horizontal range
Vi 2 sin 2 
h 
2g
R
Vi 2
sin 2
g
(1)
(2)
Using (1)/ (2), we can get
h
sin 
h
1

 tan   4 ,   tan (4h / R)
R 4 cos 
R
If h = 72.3m, R = 111 m, plug into above formula, you will get θ=69o
Then plug into (1), you can get Vi = 40.3m/s
Quiz3 (due day: march 9th, Friday)
1.You may use this way to get a bullet speed. Suppose a bullet of
mass 8.00 g is fired into a block of mass 2.50 Kg initially at rest at
the edge of a frictionless table of height 1.00m. The bullet
remains in the block, and after impact the block lands a distance
2.00 m from the bottom of the table, Determine the initial speed
of the bullet. (review perfectly inelastic collision and projectile
motion)
Solution:
Suppose the speed of bullet is Vb, bullet is shot inside the wood
block, that is perfectly inelastic collision, so momentum need be
conserved, suppose the speed of block with bullet is V1,
mb Vb = (mb +mw) V1, (1) We know the height of table is 1m, and
distance from the edge is 2m,
From the projectile motion kinematics, we can get:
V 1t  2 m
1
g ( t ) 2  1m
2
t  0.45s, V 1  4.427m / s
Then we plug in (1), get Vb = 1388m.s
Quiz 4: Rebounding pendulum ( see the example 10.9, page 286)
Mb =1Kg
In this problem, If we set mB is 1Kg
(1) What is the speed the ball A just before impact the block ? (2.4m/s)
(2) What is the speed VA and VB after collision?(-1.6m/s, 0.8m/s)
(3) What is rebounding angle θ3 ?(29.6o)
Quiz5: Figure shows a contest in which a sphere, a cylinder, and
circular hoop all of mass M and radius R( notice particle is not
taking the race). are rolling down from rest at the same instant
of time. ( please read text book P 366-367 first, then do your
own work)
(1) What is the speed of center of
mass for each of the object
when they reach the bottom
(2) What is acceleration
of the center of mass for each?
(3) Which will win the downhill
race? why
Due day: March 28, Wed.
If we choose the bottom of the ramp as the zero point of
potential energy, from energy conservation
1/ 2 Icm 2  1/ 2MVcm  Mgh,
Icm  cMR 2 c is constant
and Vcm   R
Here C is a constant that depends on the object’s geometry. C = 1 for hoop, =1/2 for cylinder
= 2/5 for sphere From above equation
Vcm 
2 gh
,
1c
Vcm  gh
for hoop,
4 gh
3
for cylinder,
g sin 
Vcm  2aS S  h / sin  , a 
1 c
2
acm  g sin  / 2 for the hoop,
acm  2 g sin  / 3 for the cylinder,
acm  5g sin  / 7 for the sphere
10 gh
for sphere
7
Continue:
S  h / sin   1/ 2acmt 2
1
t  2h /(sin   acm) 
sin 
2h(1  c)
g
1
Vcm  acmt  t  Vcm / acm 
sin 
2h(1  c)
g
Since they have the same h and θ, the greater of c, the longer of time.
Hoop’s c =1> cylinder’s c = ½, > sphere’s c = 2/5, that results in
Sphere taking shortest time, cylinder second, hoop is the last
Quiz 6:
Two equal masses are attached to identical ideal strings,
length L, next to one another. One mass is pulled 10o and
the other pulled only 5o , then release. The masses are
released simultaneously. Which mass reaches the vertical
position( θ = 0o) first? Explain your reasoning.
1. (due day: April 11th Wednesday)
Answer: They reach the equilibrium point at the same time
Reason: For the small angle pendulum, the period
T  2
L
g
If they have the same L, they will have the same period. The
maximum angle does not effect the period.
And the time reaches the equilibrium position should be T/4, since two
pendulum system have the same period, two masses should reach the
equilibrium position at same time.
Note: at equilibrium position θ(t) = 0, the function for this oscillation is
 (t )   m cos(t )  0  cos(t )  0
2

t   / 2, 
t  ,t  T / 4
T
2
( if you pull out mass to the maximum and release, the phase constant φ=0)