Problem 3
Two hockey pucks collide on a
horizontal, frictionless surface. The
velocities of both pucks before the
collision are shown in the drawing.
After they collide, puck A is moving
as shown. Assume that the mass of
puck A is twice as large as the mass
of puck B: MA = 2*MB.
BEFORE
A
40
AFTER
o
A
25 m/s
30 m/s
25
B
15 m/s
B v=?
o
a) Find the magnitude and direction of the velocity of puck B after the collision. Clearly
indicate on your drawing the angle you are using to specify the direction of puck B.
b) Is kinetic energy conserved in this collision? Justify your answer. If not, clearly indicate
whether KE is gained or lost.
Problem 4
a) A baseball-player-turned-astronaut stands on Mimas, a small moon of Saturn, and
throws a baseball with initial speed 37m/s. The mass of the baseball is 0.15kg ; the mass
19
of Mimas is 3.8x10 kg and its radius is 200km. If the astronaut throws the ball
horizontally, will it go into a circular orbit around Mimas? Explain your answer.
b) For the same situation given in part b), if the astronaut throws the ball vertically, how
high above the surface of Mimas will it rise before stopping?
Problem 5 Short Answer Questions:
You must show your work or write an explanation of your answer to get credit
for these problems.
i) A ball hits a wall and bounces off as shown. Assume that the collision is elastic. Which
vector best represents the direction of the change in momentum of the ball?
a) Circle your choice and explain it b) Not enough information to answer.
A
B
C
D
E
F
G
H
ii) A small car is moving along a straight level highway at a speed of 3v. The car hits from
behind a large truck moving in the same direction at speed v. After the collision, the car
is stuck onto the truck. During the collision, which vehicle experiences the greater
average force? Explain your answer.
a) The car
b) The truck
c) The forces are equal
d) Impossible to determine without information about the masses
2
2
+ 21 MB vBI
= 12 (2MB )(25)2 + 21 MB (30)2 = 1075MB
b) KEI = 12 MA vAI
1
1
2
2
KEF = 2 MA vAF + 2 MB vBF = 12 (2MB )(15)2 + 21 MB (40.5)2 = 1045MB
KE is lost
Problem 4
a) For a circular orbit
GMm
r2
=
mv 2
r
⇒v=
�
GM
r
r = 2 × 105 m ⇒ v = 113m/s ⇒ 37m/s is too slow
b) KEI = 12 mvI2 P EI =
−GM
r2
=
−GM
r1
+ 12 vI2 ,
−GMm
, KEF = 0
r1
vI2
1
1
1
r2 = r1 − 2GM , r2 =
⇒ height = 1.1 × 104 m
P EF =
4.73 ×
−GMm
r2
10−6 , r2
= 2.11 × 105 m
Problem 5
i)
ii) c) The forces are equal. Newton’s 3rd Law: Action/Reaction.
iii) c) At an angle not equal to 0◦ or 180◦ .
P1 = 8, P2 = 6, Pf = (2 + 3)(2) = 10,
P�1 + P�2 = P�f .
3-4-5 right triangle.
Problem 6
a) pT OT = 14, pInitial = (10)(3) − (4)(4) = 30 − 16 = 14, pF inal = (10)vA + 24
14 = 24 + 10vA , vA = −1 m/s = 1 m/s to the left
b) KEI = 12 (10)(9) + 12 (4)(16) = 45 + 32 = 77
KEF = 12 (10)(1) + 12 (4)(36) = 5 + 72 = 77
KEI = KEF , collision is elastic.
c) F = Δp
Δt , Δp = pf − pi = (4)(6) − (4)(−4) = 24 + 16 = 40
40
F = −3 = 40, 000 N = 40, 000 N to the right
10
2