55 Chapter 9, Problem 50 Bookmark Show all steps: (n :5
Step-by-step solution
Step 1 of 1 A
Answer:
< Accotding w 11: law of canservalicm of momentum, The total momentum of an isolated >
system of object remains constant. NEXT
Letusconsiduthewmandshasasys

55 Chapter 9, Problem 28F a Bookmark Show all steps: (n ;:
Step 1 of4 A
Mass of the tennis ball (m) = 0,060 kg.
Step 2 of 4 A
300
m4) 200
100
0
0.05 am
I0)
Comment
Step 3 of 4 A
(a) The impulse = afea under the F- t curve.
Area of each block =50N><0.0ls.

55 Chapter 9, Problem 40 [1 Bookmark Show all steps: (n :
Step-by-step solution
Step 1 of 2 A
Solution:
< Impulse: It is the product of average force applied and time interval. It is also equal to the >
change in momentum of the body. NEXT
Step 2 of 2 A
D

Chapter 3, Problem 31 P
D Bookmark
Show all steps: (n
I!
3
Comment
Step 4 of 6 A
The same range of 2 m can be obtained from another angle. That is,
0 = 90- [8
= 72
Comment
Step 5 of 6 A
Therefore, there will be two angles which will give the

55 Chapter 2, Problem 75613 a Bookmark Show all steps: (m- ;:
2g
_ (13.89 mls)
' 2(930 mls)
=9.84 m
~-
Step 4 of 4 A
(t)
From the part (a) we got the result as the nal velocity of the car as it hits the yound
when it is dropped from a height H is v = .f2g

Chapter 9, Problem 10
<
Step-by-step solution
Step 1 of2 A
Salado 11:
Law of conservation of momentum: In the absence of external force, the total
momentum of a system remains constant.
Step 2 of 2 A
Momentum is conserved in the absence of a net external

55 Chapter 2, Problem 7SGP a Bookmark Show all steps: m3
Step1 of4 A
APPROACH:
< By using the equation 122 = v: +2a(yyo)we can approach the solution for the given >
problem, Here vuis the initial speed of the car,vis nal speed of the car, yuis the initial

55 Chapter 2, Problem 74GP a Bookmark Show austeps: (ID 3:
Step 1 of4 A
Height above the water (11) = 16m
< Reaction time (t) = 0.20 s >
Step 2 of 4 A
We use a coordinate system with the origin at the top of the water and the upward
direction as positive.

55 Chapter 3, Problem 35P Bookmark Show all steps: (n g:
. . . l 2
We have a kinematics equation yyo = (+5!
I
Ea, +vony=0
This is a quadratic equation and the roots of equation give the value of time.
2 1
vyo i V 4 50 (-y)
1
2(5 (1,]
-8.05m/s:l: ,l(8.05m/

55 Chapter 3, Problem 33F 11 Bookmark Show all steps: (n :
Step-by-step solution
Step 1 of 2 A
APPROACH:
When a foot ball is kicked to up at an angle with the horizontal follows
< Parabolic trajectory. The total time that it travel from starting point

55 Chapter 9, Problem 6P a Bookmark Show all steps: (In. ;:
Step-by-step solution
Step 1 0M A
Use law of conservation of energy to find the velocity of the ball. Later use the formula for the
< average force to calculate the average force on the ball. >
N

Chapter 2, Problem 70GP
D Bookmark
(5.0 m/s)! =%(1.2 mils):2 [-.- from equations (1) and (2)]
1
5.0 m/s = 5(12 ails):
_ 2(5.om/s)
- 2
1.2 m/s
=8.33 5
Comment
Step Sofa A
For fugitive, time taken to reach his nal velocity is 5.0 3. But time required to cat

55 Chapter 9, Problem 6P [1 Bookmark Show all steps: (n :
Step 3 of 4 A
The average force on the ball is,
17:
Substitute (26.746 run/s); for v, (32.0 m/s)(.) for 60, ans kg for m, and 25 ms for N.
0.l45 kg(26.746 m/s)j'(32.o m/s)(i)
4
2.5 ms[lo 5)
1 ms
=(

55 Chapter 9, Problem 8P [1 Bookmark Show all steps: (n :
Step 1 of3 A
The speed of air (v) = 120 kmfh
< = [120 x %]mls >
=33,33m/s
The density of air (3):12cglm3
Area of the building (A) = (45m) (65m)
Step 20f3 A
In a time dz, all of the air within a dis

55 Chapter 2, Problem 71GP a Bookmark Show all steps: (m- ;:
Step 1 of2 A
The acceleration due to gravity on the moon is g = g the value of g on Earth,
Step 20f2 A
We assume the same initial velocity on the Moon and Earth.
.At the highest point, the final

55 Chapter 9, Problem 4P [1 Bookmark Show all steps: (n :
Step-by-step solution
Step 1 of2 A
The mass ofthe particle = m
< The force on the particle (F) = 26i 12tj >
Fis varyingfrom t= 1.0 s to t=2.0 5
Step 20f2 A
In terms of momentum, Newtons Second Law

55 Chapter 9, Problem 7P a Bookmark Show all steps: (m- ;:
Step 2 of 2 A
SOLUTION
The final velocity
v = .wtan 5
From the law of conservation of momentum, initial linear momentum is equal to nal
linear momentum
Pu. =Pm
Rocket initial velocity is u = 0 , s

55 Chapter 2, Problem 7SGP Bookmark Show all steps: @ ;:
Step 1 of3 A
APPROACH:
< By using the simple kinematics equation v2 =v:+2a(xxo)we can approach the >
solution for the given problem. Here vois the initial speed of the car, vis nal speed of the
cany

EE Chapter 9, Problem 3P a Bookmark Show all steps: (n ;:
Step-by-step solution
Step 1 of2 A
The momentum of the particle is P = 4.8t2i-8,0j-8.9Ik
NEXT
Step 2 of 2 A
In terms of momentum, Newton's Second Law can be written as
Fons FEE
d!
i 4.8-8.0j-8.9zk

55 Chapter 3, Problem 29P [1 Bookmark Show all steps: (n g:
In vertical direction
Initial height is y, = 0
Final height ofrock (y) = 75m
Acceleration is (a)= g
Time taken to reach the water is (t) = 3.05
In horizontal direction
Initial velocity of driver

55 Chapter 3, Problem 32P Bookmark Show all steps: @ ;:
Step 1 of 2 A
APPROACH:
. . . . . 1 2 A2:
< By usmg Simple kinematics equations y = y0 +vyo +ayt and V: = Twe >
can nd the initial speed of the ball.
Step 2 of 2 A
SOLUTION:
Let the origin is at the

55 Chapter 9, Problem 70 [1 Bookmark Show a slepsi (n :
Step-by-step solution
Step 1 all A
The magnitude of momentum change is the same for both the small car and the truck.
< From Newton's 3"! Law, the force exerted on the truck by the ear, FCpl' and the

55 Chapter 9, Problem 5P a Bookmark Show austeps: (ID 3:
step 20rd A
Initial momentum of the baseball is equal to the product of its mass and speed, directed along
the X-axis,
13m! = ("me )i
Substitute 145 g for mand 30 m/s for v
5" cfw_5 4%)](30 ms):
=(o

55 Chapter 9, Problem SOP Bookmark Show all steps: @ ;:
Step4of6 A
(b) The impulse given to the bullet is the area under the ft graph
= area of triangle
=[%][3x10_3s)[740N)
J=
Step 50f6 A
(1:) By integation
Impulse J = IF -dt
= cfw_rm(7m 2.3x105t)at
3x "1

55 Chapter 2, Problem 73GP a Bookmark Show all steps: (m- ;:
SOLUTION:
Initial velocity (v0) ofthe car: 100 kmlh
=[100 EJ[1000m][ In J
h 1km 3600s
= 27.78 mls
Final velocity (v)ofthe car: 0 mls
Deeeleration (a ) of the car = 30 g
= (3o)(9.so mls)
= 29

55 Chapter 9, Problem 1P Bookmark Show 51 Step5: (E- :i
Step-by-step solution
Step 1 of 2 A
dM
The rate ofpropellant gases expelled 7 = 1300 kg/s
< Speed of the gases (v) = 45,000 m/s >
Comment
Step 2 of 2 A
The force on the expelled gases calculated from

55 Chapter 3, Problem sop a Bookmark Show all steps: (m- ;:
Step 1 of 3 A
APPROACH:
< This problem depends on the concept that the acceleration due to yavity on Moon is % >
times that of Earth.
Comment
Step 2 of 3 A
SOLUTION:
Let the acceleration due to g

55 Chapter 9, Problem SOP [1 Bookmark Show a 51995: (E- :5
J = 11le
Step 5of6 A
(c) By integration
Impulse J = [F dt
= fww 2,3x10t)dt
Sun-"ls
=[740z[2'3ZIOSJzl
=[(740)[3x10-3s)_[23:105](3X10_35)2:l
=[2.22 Ns1.035 N3]
=-
Step time A
(d) We nd the mass

4/8/2015
Catapult Lab
Abstract:
In this lab, a catapult was built with shooting in three different angles. The angles are 15 degree,
30 degree and 45 degree. The catapult was built using rubber bands, strings, tapes, glue and
mainly popsicle sticks and a