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EXAM # 1 – PHYSICS 131
FALL – 2016
NAME: ____________________Key_________
ID#: __________________
Part 1 [20 Pts]
Part 1 has 5 multiple choice questions of equal weight. Circle the right answer (only one).
1. Using the given units for the variables, which one of the following expressions is correct.
T has units of s, L in m and g in m/s2.
a) T = 6.28L/g
b) T= 0.16/( Lg)
c) T2 = 6.28 Lg
d) T2 = 6.28 (L/g)
Ans d)
2. A ball is kicked at an angle of 45o with respect to ground. Which one of the followings
best describes the acceleration of the ball during this event if air resistance is neglected?
a) Acceleration is zero all the time.
b) The acceleration is 9.8 m/s2 downward all the time except that at the top it drops
to zero.
c) The acceleration is 9.8 m/s2 downward all the time.
d) The acceleration starts at 9.8 m/s2 downward but then drops to some lower value
as ball approaches the ground.
Ans c)
3. Two cars A and B travel around the same curve, car A at twice the speed of the car B.
Which of the following is true?
a) Both cars have same centripetal accelerations.
b) The centripetal acceleration of car A is double than that of car B.
c) The centripetal acceleration of car A is half than that of car B.
d) The centripetal acceleration of car A is four times that of the car B.
Ans d)
4. Complete the following statement: The maximum speed at which a car can safely
negotiate a frictionless banked curve depends on all of the following except
a) the angle of banking.
b) the radius of the curve.
c) the mass of the car.
d) All of the above.
Ans c)
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5. The kinetic energy of a car is 8 × 106 J as it travels along a horizontal road. How much
power is required to stop the car in 10 s?
a) zero watts
b) 8 × 105 W
c) 8 × 106 W
d) 8 × 107 W
Ans b)
Part 2
1. A projectile fired from a gun has initial horizontal and vertical components of velocity
equal to 30 m/s and 40 m/s, respectively. Find
a) [4 Pts] The initial velocity (speed and direction) of the projectile.
𝟒𝟒𝟒/𝒔
v0 = √𝟑𝟑𝟐 + 𝟒𝟒𝟐 = 50 m/s and 𝜽 = 𝐭𝐭𝐭−𝟏 𝟑𝟑𝟑/𝒔 = 53.1o with x-axis.
b) [6 Pts] Maximum height of the projectile.
Vertically v0= 40 m/s, v = 0 (at top), a = -9.8 m/s
2ay = v2 - v02 H= y = (40)2/(2*9.8) = 81.63 m
2. An ice skater is gliding horizontally across the ice with an initial velocity of +5.0 m/s.
The coefficient of kinetic friction between the ice and the skate blades is 0.050, and air
resistance is negligible.
a) [6Pts] What is the acceleration of skater?
(It was given for recitation)
Vertical: FN = mg
Horizontal: Net F = fk = -ma , Also we know fk = μk FN = μk mg
Therefore, μk mg = -ma => a = -μk g = - 0.49 m/s2.
Hence a = 0.49 m/s2 in backward direction
b) [4 Pts] How much distance skater travels before she finally comes to stop?
Now v0 = 5.0 m/s, v = 0 m/s, a = - 0.49 m/s2, x =?
We have 2ax = v2 - v02  x = (v2 - v02)/2a
Hence x = 25.5 m
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3. A block of ice slides freely (no air resistance, no friction) down a slippery slope. The
block starts from rest from the top of the slope, where it had a vertical height of 16 m
above the bottom of the slope and where it had a gravitational potential energy of 800 J.
a) [4 Pts] At quarter of the way down (at height of 12 m), find its kinetic and
gravitational potential energies.
PE = mgh
 600 J
while total energy is conserved = 800 J. Now it is ¾ of 800 J
The total energy is conserved = 800 J while PE we got is 600 J,
Therefore KE is 200 J.
b) [6 Pts] Find the speed of the block when it arrives at the bottom of the slop.
At top E0 = PE only = mgh
At bottom Ef = KE only = ½ mv2
Conservation of energy mgh = ½ mv2
Hence 𝒗 = �𝟐𝟐𝟐 = 17.71 m/s
Exam-1 Formula Sheet
You may use the following formulas and equations
∑F = m×a
f
µax
= µFN
FG = G
m1m2
R2
cos θ =
ha
h
sin θ =
tan θ =
ho
h
ho
ha
W=m×g
v2
ac =
r
Work = F × s × Cos (θ)
Gravitational PE = m × g × h
KE = ½ m × v2
E (total) = PE + KE
h 2 = ho2 + ha2
v = v0 + at
1
x = v0 t + at 2
2
2ax = v 2 − v02
Power = Work / Δt
G = 6.674×10-11 N.m2/kg2
REarth = 6.38×106 m
MEarth = 5.98×1024 kg
Solution of a Quadratic equation,
𝑎𝑥 2 + 𝑏𝑏 + 𝑐 = 0,
2
−𝑏 ± √(𝑏 − 4𝑎𝑎)
𝑥=
2𝑎