HW#8a
Note: numbers used in solution steps can be different from the question part. You can practice the methods in
Page 1 of 7
the solution and verify with the numbers and answers given in the question part. Or you should practice the
methods in the solution and verify your calculation with numbers in your webassign.
Problem 1, On a horizontal frictionless table, masses A and B (2 kg, 3 kg) slide to the right and left,
respectively. They have speeds of 3 m/s and 1 m/s, respectively. The two masses collide, and bounce off each
other. After the collision, they travel in opposite directions at speeds of 1.5 m/s and 2 m/s, respectively.
(Note:In this problem, please use + dir = right, - dir = left, to indicate direction.)
a.) Calculate A's momentum: before collision:
Explain why it is NOT conserved.
kg·m/s , after collision:
kg·m/s
Solution:
Speeds were given, assign directions +/- appropriately:
vAix= + 3 m/s
vBix= -1 m/s
vAfx= -1.5 m/s
vBfx= +2 m/s
pA is not conserved. It goes from +6 kg m/s before the collision to -3 kg m/s after the collision. The reason it is
not conserved is that pA is only conserved when the net force on object A equals zero. But A does feel net force
during collision (from B).
b.) Calculate B's momentum: before collision:
Explain why it is NOT conserved.
kg·m/s
after collision:
kg·m/s
pB is not conserved. Similar to part A, pB is not conserved since net force on object B is not zero. B feels a force
during collision (from A).
c.) Calculate the A+B system's momentum:
kg·m/s
before collision:
Explain why it IS conserved.
after collision:
kg·m/s
p of A+B system is conserved, since Fnet sys= 0. The total net force on the system is zero. The force between A
and B are internal forces which do not change the total moments of the system.
Use p conservation for A+B, since no net external forces on the system:
ƒ N and mg cancel
ƒ Frictionless
ƒ Collision force are internal to system
HW#8a
Calculate the kinetic energy of the A+B system:
before collision:
J
after collision:
What kind of collision is this? somewhat inelastic
Page 2 of 7
J
KE decreased. Mechanical energy is lost to heat/deformation.
This is a somewhat inelastic collision. Mechanical energy is lost to heat/deformation (but don’t stick together,
which would be perfectly inelastic). Somewhat inelastic collision does not stick together. Don’t assume elastic
and conservation of KE, when object do not stick together. Only if there is no any lose to deformation, heat,
sound, etc. you have KE unchanged and that’s called elastic collision. That’s an ideal case for extremely hard
materials. Most collisions in reality are in between----- somewhat inelastic collision.
Insight: for all kind of collision, total P initial = total P final is the only thing you can always trust.
Problem 2, In a football game, a tackle running at a constant speed of 4.5 m/s tackles a stationary receiver. The
two fly off horizontally at 2.6 m/s, before they hit the ground and begin to slow down. If the mass of the tackle
is 115kg,
a.) What is the mass of the receiver?
kg
b.) Calculate the kinetic energy of the tackle-receiver system before and after the collision.
KEsys, i =
J
KEsys, f =
J (Did the mechanical energy of the system increase, decrease, or stay the same?)
---Select---
c.) What kind of collision is this?
(Did you need to do part b to be able to answer this?)
completely inelastic
system = both people
Can say that momentum of this system is conserved since no net external force acts on it:
ƒ N and mg cancel (horizontal)
ƒ Since tackle has constant velocity before collision (see Newton #1), we know friction must somehow be
canceled by running force.
ƒ Tackle force is internal to system
(It is true that after collision, the system would immediately begin to slow down (decrease p), if friction
presumably became an external force on system as they slid along the ground. Nonetheless, our calculation can
yield the speed of both people the moment right after the collision, before they have a chance to slow down to
any net external forces.)
(A)
Total Pxi = Total Pxf, there is no y direction.
MA VAix + MB VBix = MA VAfx + MB VBfx ,
Notice that VBix =0
Also, we know that they stick together after collision so that VAfx =VBfx = Vf
So that : MA VAix = MA Vf + MBVf ,
Hence, MB = (MA VAix - MAVf )/Vf =(115*4.5-115*2.6)/2.6 = 84 kg
(B)
HW#8a
KEsys, i = ½ MAVAix2 + ½ MBVBix2 = 1/2 *115*4.52 =1164 J
KEsys, f = ½ MAVAfx2 + ½ MBVBfx2 = 1/2 *115*2.62 + 1/2 *84*2.62 =673 J
Page 3 of 7
KE is not conserved. It is decreased. Mechanical energy is lost to heat/deformation.
C) This is a completely inelastic collision. You know this from the start, since if the two people stick together; it
is by definition a completely inelastic collision. So the answer for part b should be surprising Æ mechanical
energy is always lost to heat/deformation in a collision of this kind.
Problem 3, Note: Please be careful with roundoff error in this problem.)
The drawing shows a collision between two pucks on an air-hockey
table. Puck A has a mass of 2.8 kg and is moving along the x axis with a
velocity of +5.5 m/s. It makes a collision with puck B, which has a mass
of 6.3 kg and is initially at rest. After the collision, the two pucks fly
apart with the angles shown in the drawing.
(a) Find the final speed of:
puck A:
3.38m/s
2.26m/s
puck B:
(b) Find the kinetic energy of the A+B system:
before the collision:
42.3J
after the collision:
32.2J
(c)What type of collision is this?
Solution: Again, no matter what kind of Collision it is, the momentum of this system is conserved since no net
external force acts on the system. Total Pi = Total Pf. Notice that P are vectors, they have both x and y
components. For the total P to conserve, the system must have momentum conservation in both x and y
directions:
Total Pix = Total Pfx
And
Total Piy = Total Pfy
Hence:
MA VAix + MB VBix = MA VAfx + MB VBfx ,
and
MA VAixy + MB VBiy = MA VAfy + MB VBfy ,
The rest is pure math.
VAiy = VAf sin(65)
VAix = VAf cos(65)
VBix = VBf cos(37)
VBiy = -VBf sin(37) ATTENTION that B’s contribution in y direction is negative!!!
Plug them into the 2 equation sets you have:
MA VAix + MB VBix = MA VAf cos(65) + MB VBf cos(37) ,
and
MA VAixy + MB VBiy = MA VAf sin(65) - MB VBf sin(37) , Notice the negative sign here.
2.8*5.5 + 0 = 2.8 cos(65) VAf + 6.3 cos(37) VBf ,
0 = 2.8 sin(65) VAf - 6.3 sin(37) VBf ,
and
You always have two equations and you have to solve the equation set to find VAf and VBf, there is no way to
use only one equation to find one first.
After you plug in all numbers of Mass and cos/sin, the algebra is straight forward. If you have problem solving
equation set, use Aleks or other math courses.
The second equation tells you that VAf = (6.3 sin(37) /2.8 sin(65) )VBf = 1.494 VBf,
Plug it back to the first equation you find 2.8*5.5 = 2.8 cos(65) 1.494 VBf, + 6.3 cos(37) VBf ,
Hence, VBf = 2.26m/s
HW#8a
VAf = 1.494 VBf = 3.38m/s
Page 4 of 7
Attention: The most common mistake is that you forget object B’s final velocity component in
the y direction is negative!
(b) Find the kinetic energy of the A+B system:
before the collision:
Total KE initial = ½ MAVAi2 + ½ MBVBi2 = ½ 2.8 * 5.52 + 0 = 42.3J
after the collision:
Total KE Final = ½ MAVAf2 + ½ MBVBf2 = ½ 2.8 * 3.382 + ½ 6.3 * 2.262 = 32.2J
Notice that for KE since it is scalar not vector, direction does not matter, only the total velocity magnitude
2
2
square matters. You may wonder why for object A, I didn’t use ½ MAVAfx + ½ MAVAfy to calculate the KE
2
2
2
after collision. You soon find that ½ MAVAfx + ½ MAVAfy is exactly equal to ½ MAVAf
Because VAfx2 + VAfy2 = VAf2 . For KE we never need to worry about the directions or components, just
multiply ½ M with speed square.
(c)What type of collision is this?
They didn’t stick together, not completely inelastic collision.
They lost some KE, not elastic collision. It is again somewhat inelastic collision.
Problem 4, The question part and the solution part use different numbers.
Attention: round off error in part a, for example 0.01*106 m, is much less than 1% of that answer, so the
webassign will agree with you and give you the point.
But when you use that result to do the second part, round off error of 0.01*106 m can be close or more than 1%
of the answer for the second question, and webassign may not agree with you. To avoid that, as I mentioned in
the forum, you want to keep one more significant digits when you calculate for part 2 and only round off at the
very end.
It is important to know that no matter how big the spheres and objects are, when we calculate the mass center
of the system, we can always consider the big sphere or object concentrated its own mass at its one mass
center and don’t worry about their size. Because its size and shape already determined its own mass center
position, we do not need more detailed info to find the system’s total mass center.
HW#8a
Treat the big object as a point at its own mass center is what we should do.
Page 5 of 7
Problem 5, The question part and the solution part use different numbers.
Problem 6, Four masses sit at the corners of a rectangle as shown. If mA=5 kg and mB=4 kg, and d1=3 m and
d2=6 m,
y
6m
5 kg
5 kg
x
3m
4 kg
4 kg
a.) Conceptually, where should the center of mass be in relation to the geometrical center of the rectangle
(marked with +)? Explain.
b.) Find the exact location of the center of mass.
Solution. (a) Conceptually, the center of mass should be a bit towards the larger masses.
(b) The horizontal position of the center of mass is surely at the center, because of the symmetry of
the masses. The vertical coordinate, yCM, of the center of mass is calculated as the mass-weighted
average coordinate:
1
yCM =
( ( 4 kg )( −1.5 m ) + ( 4 kg )( −1.5 m ) + ( 5 kg )( +1.5 m ) + ( 5 kg )( +1.5 m ) )
5 kg + 5 kg + 4 kg + 4 kg
1
=
( 3 kg m )
18 kg
= 0.16667 m
HW#8a
Page 6 of 7
Problem 7, The minute and hour hands of a clock have a common axis of rotation and equal mass. The minute
hand is long, thin, and uniform; the hour hand is short, thick, and uniform. Is the moment of inertia of the
minute hand greater than, less than, or equal to the moment of inertia of the hour hand?
greater than
less than
equal to
Solution or Explanation
The long, thin minute hand (with mass far from the axis of rotation) has the greater moment of inertia.
Problem 8,
Problem 9,
Problem 11 , A rigid object rotates about a fixed axis. Do all points on the object have the same angular speed?
yes
no
Do all points on the object have the same linear speed?
yes
no
HW#8a
Key: All points on the rigid object have the same angular speed. Because they cover the same amountPage
of angle
7 of 7
for the same amount of time. Not all points have the same linear speed, however. The farther a given point is
from the axis of rotation the greater its linear speed. Further points cover longer distance within the same
amount of time in rotation.
Problem 12, Can you drive your car in such a way that your tangential acceleration is zero while at the same
time your centripetal acceleration is nonzero?
yes
no
Key: This is the situation whenever you drive in a circular path with constant speed.
When your angular velocity is not changing, your tangential velocity is not changing; you have no angular
acceleration, nor tangential acceleration (no speed change in the tangential direction). But as long as you are
moving in a circle, you keep change your velocity directions and you always need the centripetal acceleration
(which is always pointing to the center) to keep you on the circle track.