Honors Physics
WMMHS
Mr. Quinn
2/9/11
NAME ___________________________________
** Chapter 7 Test (100 Points) **
Part 1
Multiple Choice (20 Points)
DIRECTIONS: Choose the best answer to complete the following questions.
1. _____D_____ An impulse applied to an object will proportionally change its:
a.)
b.)
c.)
d.)
Kinetic energy
Acceleration
Position
Velocity
2. _____B_____ Net force can be described as:
a.)
b.)
c.)
d.)
p * t
p / t
t / p
p – t
3. ____B______ A box moving 3.00 m/s, East, is pushed until its velocity is 3.00 m/s, North.
The direction of the average net force applied to the box is:
a.)
b.)
c.)
d.)
North of East
North of West
South of East
South of West
4. _____B_____ Which of the following is TRUE for all collisions?
a.)
b.)
c.)
d.)
Kinetic energy is conserved
Momentum is conserved
v01 + vF1 = v02 + vF2
None of these are always applicable.
5. _____D_____ Which of these examples is the best representation of an elastic collision?
a.)
b.)
c.)
d.)
Two football players colliding during a tackle
A boy jumping on a skateboard
An unmanned test car crashing into a rigid wall
A bowling ball colliding into a bowling pin
6. _____D_____ Car A (m = 1000 kg), initially travelling eastwardly, collides with Car B (m=
2000 kg), initially travelling southwardly with equal speed as Car A, in a perfectly inelastic
collision. The final momentum of the combined mass is:
a.)
b.)
c.)
d.)
South
Directly SE
Slightly South of East
Slightly East of South
7. ____C_____ A wooden baseball bat is illustrated below. Its center of mass is located:
a.)
b.)
c.)
d.)
In the center of its total length
At its end
Slightly toward the thick barrel from its center
Slightly toward the thin handle from its center
8. _____D_____ A car (m = 1000 kg) and a truck (m = 3000 kg) are 40 meters apart. The center
of mass of this two vehicle system is:
a.)
b.)
c.)
d.)
10 meters from the car
20 meters from the car
27 meters from the car
30 meters from the car
9. _____B_____ A heavy bowling ball hits a light bowling pin. During the collision, the force
that the ball applies to the pin is:
a.)
b.)
c.)
Greater than the force that the pin applies to the ball
Equal to the force that the pin applies to the ball
Less than the force that the pin applies to the ball
10. ____C______ A heavy bowling ball hits a light bowling pin. During the collision, the
acceleration of the ball is:
a.)
b.)
c.)
Greater than the acceleration of the pin
Equal to the acceleration of the pin
Less than the acceleration of the pin
DIRECTIONS FOR PARTS 2-5: Complete the following questions. A relevant equations sheet
is attached to the back of this test for your reference. Be sure to use PROPER UNITS, include
the correct number of SIGNIFICANT FIGURES, and SHOW ALL OF YOUR WORK!
Part 2
Impulse (20 Points)
A 5.00 kg box, initially at rest, is forced variably over a 5.00 s time interval as
demonstrated through the Force vs. Time graph below.
a.)
What is the impulse delivered to the box over the 5.00 s time interval?
Impulse = Area Under Curve = (1/2) (2) (20) + (2) (20) + (1/2) (1) (20) = 70 N*s, forward
b.)
What is the velocity of the box after the 5.00 s time interval?
mvF = mv0 = Impulse
(5) (vF) – 0 = 70
vF = 14 m/s, forward
c.)
What is the average force applied to the box during this time interval?
Favg = Impulse / Dt = (70) / (5) = 14 N, forward
Part 3
Collisions (20 Points)
A ballistic pendulum is a device that can be used to calculate the speed of a bullet fired
from a gun. A bullet is fired into a block of wood hanging from a pendulum arm. The bullet
embeds itself in the wood, and the combined mass swings vertically. Measuring the change in
height of the pendulum will yield a calculated bullet velocity before it hits the block of wood.
A 5.0 g bullet is fired into a 4.00 kg block of wood hanging from a 2.00 meter pendulum
arm, as shown above. The combined mass swings to a maximum angle of 25° to the vertical as
shown.
a.)
What is the increase in gravitational potential energy of the bullet-wood system
after the collision?
hf = L – L cos q = 2 – 2 cos 25 = 0.187 m
Dug = mghf = (4.005) (9.8) (0.187) = 7.34 J
b.)
Assuming no loss of mechanical energy due to friction, what is the velocity of the
wood-bullet system after the collision?
K0 = UF = 7.34 J
7.34 = (1/2) (4.005) (vo)2
v0 = 1.91 m/s
c.)
What is the velocity of the bullet before it hits the block of wood?
M1v01 + m2v02 = (m1 + m2) vF
(0.005) (v01) + 0 = (4.005) (1.91)
V01 = 1530 m/s
Part 4
Collisions (20 Points)
Lisa is playing shuffleboard in gym class. One of her attempts knocks an opponent’s disk
out of the scoring sector, as shown in the diagram below.
vFA = 3.46 m/s, 30 degrees above +x
A
V0A = 4 m/s, +x
B
vFB = ?
All shuffleboard disks are of equal mass. Lisa’s disk, travelling forward with velocity
4.00 m/s, hits an opponent’s disk in a glancing collision. As a result, Lisa’s puck is deflected 30°
from its original path with a final speed of 3.46 m/s as illustrated.
a.)
Apply the Law of Conservation of Momentum to the x- and y-components to
solve for the velocity (magnitude and direction) of the target disk after the
collision.
X:
m1vo1x + m2v02x = m1vF1x + m2vF2x
(4) + (0) = (3.46) (cos 30) + vF2x
vF2x = 1.00 m/s
 masses equal, so they CANCEL
Y:
m1v01y + m2v02y = m1vF1y + m2vF2y
0 + 0 = (3.46) (sin 30) – vF2y
vF2y = -1.73 m/s
 masses cancel again
 y-component velocities opposite in dir.
vf2 = 12 + 1.732
b.)
vF = 2.00 m/s
q = tan-1 (1.73/1) = 60 degrees below forward
Determine if this collision is elastic through calculation.
K0A + K0B = KFA + KFB
 If this equation is true, then the collision is elastic.
2
2
(1/2) (m) (4) + 0 = (1/2) (m) (3.46) + (1/2) (m) (2)2
 masses cancel again
8=6+2
 This is true, so the collision is ELASTIC.
Part 5
Collisions (20 Points)
A bowling ball collides elastically and head-on with a bowling pin. The 7.00 kg bowling
ball, moving with initial velocity of 3.50 m/s, collides with the 1.50 kg pin, initially at rest.
a.)
Find the velocity of the ball and pin after the collision.
V01 + vF1 = v02 + vF2
3.50 + vF1 = 0 + vF2
vF2 = vF1 + 3.50
M1vo1 + m2v02 = m1vF1 + m2vF2
(7) (3.50) + 0 = (7) (vF1) + (1.5) (vF1 + 3.50)
24.5 = (8.5) (vF1) + 5.25
vF1 = 2.26 m/s
vF2 = 2.26 + 3.50 = 5.76 m/s
b.)
If the collision lasts for 0.0010 seconds, find the average force applied to the pin
during the collision.
Favg = Dp / Dt = [(1.5) (5.76) – 0] / (0.0010) = 8640 N
c.)
Is the average force applied to the pin you calculated in part (b) EQUAL TO,
LESS THAN, or GREATER THAN the average force applied to the ball during
the collision? Explain your answer in 1-2 sentences or through calculation.
Forces are action-reaction pairs, yo. EQUAL!!
Relevant Equations Sheet
Momentum:
p = mv
p = F t
xcm 
x1m1  x2 m2  x3 m3  ...
M
Conservation of Momentum:
m1vo1 + m2vo2 = m1vF1 + m2v2F
m1vo1 + m2vo2 = (m1 + m2) vF
vo1 + vF1 = vo2 + vF2
(Perfectly inelastic collision)
(Elastic head-on collision)
Work:
W = Fd cos
Energy:
K = ½ mv2
Work-Kinetic Energy Theorem:
WNET = K
Newton’s 2nd Law of Motion and other force stuff:
F = ma
Fg = mg
Ff =  FN
Uniform Accelerated Motion:
#1
vF = v0 + at
#2
x = v0t + ½ at2
#3
vF2 = v02 + 2ax