Physics 211 Test 2 Practice
NAME:
SECTION:
Physics 211 Test 2 Practice
August 31, 2011
Multiple choice questions 2 points each.
1. Suppose several forces are acting on a mass m. The F~ in F~ = m~a refers to
(a) any particular one of the forces.
(b) the vector sum of all the forces.
(c) the force of friction acting upon the mass.
(d) the arithmetic sum of all the forces.
(e) the arithmetic mean of all the forces.
2. A 4 kg mass moving with a speed 2m/s, and a 2kg mass moving with a speed of 4m/s,
are gliding over a horizontal frictionless surface. Both objects encounter the same
horizontal force, which directly opposes their motion, and are brought to rest by it.
Which statement correctly describes the situation?
(a) Both masses travel the same distance before stopping.
(b) the 2-kg mass travels twice as far as the 4-kg mass before stopping.
(c) the 2-kg mass travels farther, but not necessarily twice as far.
(d) the 4-kg mass travels twice as far as the 2-kg mass before stopping.
(e) the 4-kg mass losses more kinetic energy than the 2-kg mass.
3. Consider two unequal masses, M and m. Which of the following statements is false?
(a) It is possible for the center of mass to lie within one of the objects.
(b) The center of mass is closer to the larger mass.
(c) The center of mass lies on the line joining the centers of each mass.
(d) If a uniform rod of mass m were to join the two masses, this would not
alter the position of the center of mass of the system without the rod
present.
4. In an elastic collision, the final total momentum is
(a) more than the initial momentum.
(b) less than the initial momentum.
(c) the same as the initial momentum.
(d) insufficient information to answer.
C W Fay IV: Spring2011
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Physics 211 Test 2 Practice
NAME:
SECTION:
5. You and your friend want to go to the top of the eiffel Tower. Your friend takes the
elevator straight up. You decide to walk up the spiral stairway, taking longer to do so.
Compare the gravitational potential energy (U) of you and your friend, after you both
reach the top, assuming that you and your friend have equal masses.
(a) It is impossible to tell, since the times and the distances are unknown.
(b) Your friend’s U is greater than your U, because she got to the top faster.
(c) Both of you have the same amount of potiential energy.
(d) Your U is greater than your friend’s U, because you traveled a greater distance in
getting to the top.
6. (5 points) An auto is costing on a level road. It weighs 10.kN. How much work is done
by gravity as it moves horizontally 150. meters? Solution: No work is done as the
force and the movement are perpendicular.
7. (5 points) Describe mechanical efficiency. Solution: The efficency is the ratio of the
work derived from a system over the energy put into a system.
8. (15 points) To get off a frozen, frictionless lake, a 68.0kg person takes off a 0.100kg
shoe and throws it horizontally, directly away from the shore with a speed of 2.00m/s.
If the person is 5.00m from the shore, how long does he take to reach it?
Given: mm = 68.0kg ms = 0.100kg vs = 2.00m/s ∆x = 5.00m t =?
Using the conservation of momentum,
pi = pf
0 = pman − pshoe
pman = pshoe
m1 vm = ms vs
ms vs
= 2.94 × 10−3 m/s
vm =
mm
(1)
(2)
(3)
(4)
(5)
(6)
since there is no acceleration in the x-direction (frictionless)
Given: ax = 0m/s2 v0 = vm
x = x0 + v0 t + 1/2ax t2
∆x = v0 t
∆x
t =
vm
mm ∆x
=
= 1.70 × 103 s
ms vs
(7)
(8)
(9)
(10)
9. (5 points) A 10g bullet moving horizontally at 400m/s penetrates a 3.0kg wood block
resting on a horizontal surface. If the bullet slows down to 300m/s after emerging from
the block, what is the speed of the block immediately after the bullet emerges?
C W Fay IV: Spring2011
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Physics 211 Test 2 Practice
v =?
Solution: Given: m1 = 10g
NAME:
m2 = 3.0kg
SECTION:
v1 = 400m/s
v1f = 300m/s
p1 = p 2
m1 v1 = m1 v1f + m2 v2f
m1 v1 − m1 v1f
v2f =
= 0.33m/s
m2
(11)
(12)
(13)
10. A 5.0kg block sits on an plane with a rough surface that is inclined at 30◦ .
(a) (5 points) Draw a free-body diagram for the block.
(b) (10 points) if the block is in equilibrium, what is coefficient of static friction?
µs =
Solution
given: m = 5.0kg θ = 30◦
use Newton’s second law and the first condition of equilibrium (a = 0m/s2 )
X
X
Fk = mg sin θ − f = 0
F⊥ = FN − mg cos θ
f
FN
f = µFN
(14)
=
=
=
=
0
mg sin θ
mg cos θ
µmg cos θ = mg sin θ
mg sin θ
= tan θ = 0.58
µ =
mg cos θ
(15)
(16)
(17)
(18)
(19)
11. A collision occurs between a mass of m1 = 1.00kg and a mass m2 = 2.00kg, before the
collision the masses have velocities of v10 = 2.00m/s2 and v20 = −0.500m/s2 , if after
the collision the first mass has a velocity of v1 = −1.00m/s2 .
(a) (10 points) After the collision, what is the final velocity of the second mass? v2 =
(b) (5 points) Is this collision perfectly elastic? Why or why not?
SOLUTION:
m1 = 1.00kg
given:
v1 = −1.00m/s2
a)
m2 = 2.00kg
v10 = 2.00m/s2
v20 = −0.500m/s2
m1 v10 + m2 v20 = m1 v1 + m2 v2
m1
(v10 − v1 ) + v20
v2 =
m2
1
v2 =
(2 − (−1)) + (−0.5) = 1.00m/s
2
C W Fay IV: Spring2011
(20)
(21)
(22)
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Physics 211 Test 2 Practice
NAME:
b) We have to answer is energy conserved?
SECTION:
1
1
2
2
+ m2 v20
= 2.25J
m1 v10
2
2
1
1
=
m1 v12 + m2 v22 = 1.5J
2
2
Ei =
(23)
Ef
(24)
No energy is not conserved, so the collision is not perfectly elastic.
12. A 20.0g ball falls on a table, If the initial velocity of the ball is 3.00m/s and after it
hits the table it has a velocity of 2.50m/s upward. The ball is in contact with the table
for 0.00100s.
(a) (10 Points) what is the force on the table? F =
(b) (5 Points) What is the efficiency of this process? ε =
Solution:
Given: m = 20.0kg
v0 = 3.00m/s
vf = 2.50m/s
t = 0.00100s
(a) what is the force on the table? F =
F = ma = m
∆v
(−3.00m/s) − (2.50m/s)
= (0.020kg)
= 110N
∆t
0.00100s
(25)
(b) (extra credit) What is the efficiency of this process? ε =
Wout
∆K
=
Ein
Ein
1 2
Wout =
mv − 0
2 2
1 2
Ein =
mv
2 1
1
mv 2
v2
ε = 21 22 = 22 = .694 = 69.4%
v1
mv1
2
ε =
(26)
(27)
(28)
(29)
13. (10 points) A ball of mass 0.500kg is thrown vertically with velocity 20.0m/s. what is
the potential energy after the speed has dropped to half of its initial value? U =?
SOLUTION:
The potential energy when the speed has dropped by half is given by the total energy
minusthe new kinetic energy,
1 v
1
1 v
3
U = ET − m( )2 = mv 2 − m( )2 = mv 2
2 2
2
2 2
8
3
=
(0.500kg)(20.0m/s)2 = 75J
8
(30)
(31)
14. (5 points) A loaded Boeing 747 jumbo jet has a mass of 2.0 × 105 kg. What net force
is required to give the plane an acceleration of 4.0m/s2 down the runway for takeoff?
C W Fay IV: Spring2011
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Physics 211 Test 2 Practice
F~ =
SOLUTION:
given: m = 2.0 × 105 kg
NAME:
a = 4.0m/s2
F = ma = 8.0 × 105 N
C W Fay IV: Spring2011
SECTION:
(32)
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