Physics 101 Fall 2007: Pledged Problems 9 – Momentum & Rotational Kinematics
Time allowed: 2 hours in one sitting
Due:
Monday, November 12 at 5PM in the box marked Phys101/102 in the Physics Lounge
You may use your own textbook, your notes and a non-programmed calculator. For the purposes of this problem set, you may also
use the online solutions to the corresponding suggested problems. You should consult no other help. Please follow the standard
format:
• Write legibly on one side of 8-1/2” white or lightly tinted paper.
• Staple all sheets (including this one) together in the upper left corner
• Make one vertical fold.
• On the outside (staple side up) on successive lines
– PRINT your last name in CAPITAL letters.
– PRINT your first name.
– Print the phrase ‘Pledged Problems 9’ and the due date.
– Print the times at which you started and finished the problems.
– Write and sign the Pledge, with the understanding you may consult the resources described above.
1. A spaceship in deep space (far from any other massive bodies) initially at rest explodes, breaking into three
pieces. Two pieces, having equal mass, fly off perpendicular to one another with the same speed of 30 m/s.
The third piece has three times the mass of each other piece.
Solution
(a) What is the direction and magnitude of its velocity immediately after the explosion?
I will assume that a piece of mass m flies off horizontally which implies that the other mass m flies
off vertically. I will choose one piece to fly off in the positive x direction and the other mass m piece
to fly off in the negative y direction. To find the direction and magnitude of the third piece, consider
momentum conservation.
∆px = 0
(1)
∆py = 0
(2)
m v0 − 3 m v0,x = 0
(3)
⇒v0,x =
1
v0 = 10m/s (To the left.)
3
m v0 − 3 m v3,y = 0
⇒v3,y =
v3 =
1
v0 = 10m/s (Up.)
3
√
200
~v3 = 10 m/s (−ı̂ + ̂)
(4)
(5)
(6)
(7)
(8)
(b) How much energy was released in the explosion?
Compare the kinetic energy of the system immediately before and after the explosion.
∆KE = KEf − KE0
∆KE = KEf = 400 m
Where m represents
1
5
(9)
(10)
the total mass of the spaceship.
2. A stream of glass beads, each with a mass of 0.5 g, comes out of a horizontal tube at a rate of 100 per
second. The beads fall a vertical distance of 0.5 m to a balance pan and bounce back to their original
height as shown in the figure below. How much mass must be placed in the other pan of the balance to
keep the pointer at zero?
Solution
To determine the mass that must be placed in the other pan of the balance, we need to determine the
average force exerted on the pan by the falling stream of glass beads. We can find the average force exerted
on one glass bead by the pan, and use Newton’s third law to find the force we need.
~ ave = ∆~p
F
∆t
Fy,ave =
∆py
∆t
(11)
(12)
∆py = py,f − py,0 = 2 py,0
(13)
py,0 = mB vy,0
(14)
By using energy conservation on a glass bead, the speed of a glass bead immediately before it collides with
the pan can be determined.
∆KE + ∆U = 0.
⇒vy,0 =
p
2 g h = 3.13m/s
⇒∆py = 3.13 × 10−3 kg m/s
∆t =
1
s
100
Fy,ave =
∆py
= 3.13 × 10−1 N
∆t
(15)
(16)
(17)
(18)
(19)
Since this is the average force exerted on the glass beads by the pan, Newton’s 3rd law ensures that the
glass beads exert a force of equal magnitude on the pan (down). This means that we need the weight on
the other pan to be 3.13 × 10−1 N .
W = 3.13 × 10−1 N = m g
(20)
⇒ m = 3.2 × 10−2 kg = 32g
(21)