Physics 102 Final Exam Review Problems

Review Problems, Physics 102, Final Exam, Fall 2023

Problem 1: Projectile Motion

A stone is thrown with an initial velocity of $50 m/s$ at an angle of $35^\circ$ .
Find the horizontal and vertical components of its velocity 4 seconds later.
Also, determine how much later it hits the ground.

Solution:

Horizontal component of velocity (vx) remains constant:
$vx = v0 \cos(\theta) = 50 \cos(35^\circ) \approx 41 m/s$
Vertical component of velocity (vy) changes due to gravity:
$vy = v0 \sin(\theta) - gt = 50 \sin(35^\circ) - 9.8(4) \approx 28.67 - 39.2 \approx -10.5 m/s$
*Therefore, the horizontal component is approximately $41 m/s$ and the vertical component is approximately $-10.5 m/s$ (Option E).
To find when it hits the ground, set the vertical displacement (sy) to 0:
$sy = v0t + \frac{1}{2}at^2$
$0 = 28.67t - 4.9t^2$
$0 = t(28.67 - 4.9t)$
Solve for t:
$t = 0$ (initial time) or $4.9t = 28.67$
$t = \frac{28.67}{4.9} \approx 5.85 \text{ seconds}$

Problem 2: Vector Magnitude

Given two vectors A and B (shown in a figure).
Find the magnitude of $3A + 6B$ .
Possible answers: A) 6, B) 12, C) 18, D) 24, E) 30

Problem 3: Significant Figures

What is the difference between $103.5$ and $102.24$ written with the correct number of significant figures?
Possible answers: A) 1, B) 1.3, C) 1.26, D) 1.260, E) 1.2600

Solution:

$103.5 - 102.24 = 1.26$
Since $103.5$ has one decimal place, the answer should also have one decimal place.
Therefore, the answer with the correct number of significant figures is $1.3$ (Option B).

Problem 4: Inclined Plane with Friction

A 2-kg block slides down an inclined plane with constant velocity. Angle of incline is $35^\circ$ .
Find the coefficient of kinetic friction.
Possible answers: A) 0.2, B) 0.5, C) 0.7, D) 0.8, E) 0.9

Problem 5: Tension Force on a Crate

A 38-kg crate is pulled along a horizontal, frictionless floor by a cord that exerts a 180-N tension force at an angle $\\theta = 30^\circ$ .
Find the normal force on the block.
Possible answers: A) 110 N, B) 146 N, C) 185 N, D) 211 N, E) 282 N

Problem 6: Block Acceleration

A 5-kg block is pulled across a horizontal plane.
A pulling force F has a magnitude of 24 N and makes an angle of $28^\circ$ with the horizontal.
An 8 N friction force opposes its motion.
Find the block’s acceleration.
Possible answers: a. 0.55 m/s2, b. 2.64 m/s2, c. 4.29 m/s2, d. 6.75 m/s2, e. 8.00 m/s2

Problem 7: Car Braking

A car has a mass of 1200 kg and is moving at 60 mi/h.
The driver applies the brakes, resulting in a net backward force of 1800 N. The brakes are held for 10 s.
How far does the car travel during this period?
Possible answers: a. 193 m, b. 175 m, c. 148 m, d. 120 m, e. 75 m

Problem 8: Gravitational Force

Two small objects, with masses m and M, are originally a distance r apart, and the magnitude of the gravitational force on each one is F.
The masses are changed to 2m and 2M, and the distance is changed to 4r. What is the magnitude of the new gravitational force?
Possible answers: A) F/16, B) F/4, C) 16F, D) 4F, E) F/2

Solution:

The gravitational force is given by:
$F = G \frac{mM}{r^2}$
The new force F' is given by:
$F' = G \frac{(2m)(2M)}{(4r)^2} = G \frac{4mM}{16r^2} = \frac{1}{4} G \frac{mM}{r^2} = \frac{1}{4}F$
Therefore, the new gravitational force is $F/4$ (Option B).

Problem 9: Impulse

A 1.0-kg ball moving at 2.0 m/s perpendicular to a wall rebounds perpendicular from the wall at 1.5 m/s.
Find the impulse of the ball.
Possible answers: A) zero, B) 0.5 N·s away from the wall, C) 0.5 N·s toward the wall, D) 3.5 N·s away from wall, E) 3.5 N·s toward the wall

Solution:

Impulse is the change in momentum:
$J = \Delta p = m(vf - vi)$
Assuming the initial velocity is towards the wall (positive direction):
$J = 1.0 \times (-1.5 - 2.0) = -3.5 N \cdot s$
The negative sign indicates that the impulse is away from the wall. Thus, the impulse is 3.5 N·s away from the wall (Option D).

Problem 10: Inelastic Collision

A 4-kg object moves to the right at 8.0 m/s. It collides with a 6-kg object moving at 5 m/s to the left. The collision is perfectly inelastic.
What is the speed of the 4 kg object after the collision?
Possible answers: A) 0 m/s, B) 0.2 m/s, C) 4.2 m/s, D) 6.2 m/s, E) 8.2 m/s

Problem 11: Ferris Wheel Deceleration

A Ferris wheel rotating at 20 rad/s decelerates with a constant acceleration of 5 rad/s2. How many revolutions will it make before coming to rest?
Possible answers: A) 4.0, B) 2.8, C) 6.4, D) 3.2, E) 1.5

Solution:

Using $\omegaf^2 = \omegai^2 + 2 \alpha \theta$ where $\omegaf = 0$ , $\omegai = 20 rad/s$ and $\alpha = -5 rad/s^2$
$0 = 20^2 + 2(-5)\theta$
$10\theta = 400$
$\theta = 40 rad$
To find number of revolutions, knowing that $2\pi$ radians is one revolution, Number of revolutions $= \frac{40}{2\pi} = 6.366 \approx 6.4$
The answer is C

Problem 12: Unwinding Wheel

A wheel of radius 2 cm has a 4-m cord wrapped around its periphery. Starting from rest, the wheel is given a constant angular acceleration of 1 rad/s2. The cord will unwind in
Possible answers: A) 125 s, B) 85 s, C) 66 s, D) 20 s, E) 15 s

Solution:

Length of cord unwound S=R $\theta$
Therefore $\theta = S/R = 4m/0.02m = 200 radians$
Using $\theta = \omega0 t + (1/2)\alpha t^2$ and knowing that $\omega0 = 0$ and $\alpha= 1 rad/s^2$
$200 = (1/2)(1)t^2$
$t^2 = 400$
$t= 20 s$

Problem 13: Ceiling Fan Inertia

When a ceiling fan rotating with an angular speed of 2.0 rev/s is turned off, a frictional torque of 0.98 Nm slows it to a stop in 4.5 s. What is the moment of inertia of the fan?
Possible answers: A) 0.35 kgm2 B) 0.55 kgm2 C) 0.85 kgm2 D) 1.35 kgm2 E) 2.35 kgm2

Solution:

First convert $\omega$ from rev/s to rad/s by multiplying by $2\pi$ , so $\omega = 4\pi \approx 12.567 rad/s$
Using $\tau = I\alpha$ Torque equals moment of inertia times angular acceleration.
We can determine angular acceleration $\alpha$ using $\omegaf = \omegai + \alpha t$ knowing $\omegaf = 0$ , $\omegai = 12.567 rad/s$ , and $t = 4.5 s$
Rearranging gets us $\alpha = (\omegaf - \omegai)/t = (0-12.567)/4.5 = -2.793 rad/s^2$
Finally, substituting into $\tau = I\alpha$ Torque equals moment of inertia times angular acceleration.
$0.98 = I * -2.793$
Isolate I by dividing both sides by -2.793.
$I=0.35 kgm^2$
The answer is A

Problem 14: Grinding Wheel Torque

An electric motor can accelerate a 2.0-kg grinding wheel in a form of a solid disk of radius 0.20 m from rest to 700 rev/min in 14 s. Find the torque generated by the motor.
Possible answers: A) 0.07 N.m, B) 0.11 N.m, C) 0.21 N.m, D) 0.33 N.m, E) 0.46 N.m

Problem 15: Figure Skater Spin

A figure skater goes into a spin, starting with her arms up and close to her body as shown. When she extends her arms horizontally sometime later the angular speed remains constant or changes and if changes how?
Answer: her angular velocity decreases.

Problem 16: Pizza and Cheese

A pizza together with its pan is rotating about a vertical axis through its center. Its rotational inertia $I = 2.00 kg.m^2$ and radius $r = 50.0 cm$ . The initial angular speed is 4.00 rad/s. A 1.00 kg chunk of Mozzarella cheese is initially at rest over the disk. It drops vertically onto the pizza from above and sticks to the edge. What is the angular speed of the pizza after the cheese becomes stuck to it?
Answer: 3.56 rad/s

Solution:

Using conservation of angular momentum principle, $Ii \omegai = If \omegaf$
Since cheese will stick to the edge ( $r = 0.5 m$ ), the final inertia is $If = Ii \omega_i + mr^2 = (2 kg.m^2) + (1kg)(0.5m)^2 = 2.25 kg.m^2$
Therefore, $\omegaf = (Ii \omegai)/If = (2 kg.m^2 * 4 rad/s)/(2.25 kg.m^2) = 3.56 rad/s$

Problem 17: Tension and Force on a Strut

Find (a) the tension T in the horizontal cable and (b) the magnitude and direction of the force exerted on the strut by the pivot in the arrangement in figure below. Let W be the weight of the suspended crate full of priceless art objects. The strut is uniform and also has weight W.
Answer: (a) T= 2.60W; (b) F= 3.27W and $\theta=37.6^o$

Problem 18: Tire Deceleration

A tire on a car is turning initially at 500 radians/s. The car brakes and slows to a stop in 20 s. What is the acceleration in Radians/s2?
Possible answers: a. -500, b. 0, c. 25, d. 700, e. -25
Using \omega = \omega_0 + at
We have \omega = 0 rad/s final, \omega_0 = 500 rad/s, and t=20s.
Substituting into the equation: 0 = 500 + a(20)
Solving for a: a = -25 rad/s^2
The answer is E. -25

Problem 19: Work to Accelerate a Car

How much work is required to accelerate a 1000 kg car from 20 m/s to 30 m/s?
A) 0.5x10^5 J, B) 1.2x10^5 J, C) 2.5x10^5 J, D) 3.8x10^5 J, E) 4.5x10^5 J

Solution:

Work done is equal to change in kinetic energy.
Work = KEfinal - KEinitial
Work = 0.5 * m * (vfinal)^2 - 0.5 * m * (vinitial)^2
Work = 0.5 * 1000 * (30)^2 - 0.5 * 1000 * (20)^2
Work = 500 * 900 - 500 * 400
Work = 450000 - 200000
Work = 250000 J = 2.5 x 10^5 J
The answer is C.

Problem 20: Car on an Incline with Friction

A 1600 – kg car starts from rest at the top of the long driveway that is inclined at 15^0 with the horizontal. If the average friction force of 1500 N impedes the motion, what is the car’s acceleration?
a. 0.76 m/s^2, b. 1.05 m/s^2, c. 1.87 m/s^2, d. 1.60 m/s^2, e. 2.75 m/s^2

Solution:

Net Force = Force due to gravity - Friction.
Force due to gravity = mgsin(15)
Force due to gravity = 1600 * 9.8 * sin(15) = 4057.7 N
Net Force = 4057.7 - 1500 = 2557.7 N
Acceleration = Net Force / mass
Acceleration = 2557.7 / 1600 = 1.60 m/s^2
The correct answer is d) 1.60 m/s^2

Problem 21: Automobile Acceleration

An automobile traveling along a straight road increases its speed from 30.0 m/s to 50.0 m/s in a distance of 180 m. If the acceleration is constant, how much time elapses while the auto moves this distance?
a. 6.00 s, b. 4.50 s, c. 3.60 s, d. 4.00 s, e. 9.00 s

Solution:

vf = vi + at
x = v_it + 0.5a*t^2
vf^2 = vi^2 + 2ax
From
vf^2 = vi^2 + 2ax
50^2 = 30^2 + 2a180
2500 = 900 + 360*a
a = 4.44 m/s^2
Substituting our result into vf = vi + at
50 = 30 + 4.44*t
t = 4.5 s

Problem 22: Car Acceleration

A car moving initially at 36 km/h, travels 174 m in 8 s along a straight line with constant acceleration. The acceleration of the car is:
A) 0. 0 m/s2 B) 0.5 m/s2 C) 1.0 m/s2 D) 3. 0 m/s2 E) 6. 0 m/s2

Solution:

Convert 36 km/h to m/s: 36 * (1000/3600) = 10m/s
Using Formula distance, $x = v_0t + (1/2)a*t^2$
174 = 108 + 0.5a*8^2
174= 80 + 32*a
94 = 32*a
a = Acceleration = 2.9375 which is approximately 3.0 m/s^2

Problem 23: Airplane Propeller

An airplane propeller starts to turn from rest and speeds up to 2 radians/s after turning 6 Radians. How long does it take, in s .
a. 6, b. 4, c. 3, d. 2, e. 1

Solution:

\omegaf^2 = \omegai^2 + 2a\theta
2^2= 0+2a6
a= 4/12 =1/3
Using equation
\omegaf= \omegai + a*t
2= 0+ 1/3 *t
t= 6 s

Problem 24: Crate on Incline

A 20 N force F, parallel to the incline is required to push a certain crate at acceleration of 1.6 m/s2 up an incline that is $\\Phi= 28^o$ above the horizontal. What is the mass of the crate?
A) 2.31kg, B) 3.49kg, C) 3.22kg, D) 5.84kg, E) 6.04 kg

Solution:

F - mgsin(Φ) = ma
20 - m9.8sin(28) = m*1.6
20 = m(1.6 + 9.8*sin(28))
m = 20 / (1.6 + 9.8*sin(28))
m = 3.22 kg

Problem 25: Block on Inclined Plane with Friction

A 1.0-kg block is pushed up a rough 22^0incline plane by a force of F=7.0 N acting parallel to the incline. The acceleration of the block is 1.4 m/s^2 up the incline. Determine the magnitude of the force of friction acting on the block.
A) 1.9 N, B) 2.2 N, C) 1.3 N, D) 1.6 N, E) 3.3N

Solution:

F - F_friction - mgsin(θ) = ma
7 - F_friction - 19.8sin(22) = 1*1.4
7 - F_friction - 3.67 = 1.4
F_friction = 7 - 3.67 - 1.4
F_friction = 1.93 N Approximately 1.9 N

Problem 26: Ball and Spring

A 5-kg ball is dropped from a height y0 above the top of a vertical spring whose spring constant is 2000 N/m. The spring is compressed 0.4 m. The height y0 is closest to
Answer: 2.86 m

Problem 27: Block, Spring, and Friction

A 1.5-kg block sliding on a rough horizontal surface is attached to one end of a horizontal spring (k = 200 N/m) which has its other end fixed. If this system is displaced 20 cm horizontally from the equilibrium position and released from rest, the block first reaches the equilibrium position with a speed of 2.0 m/s. What is the coefficient of kinetic friction between the block and the horizontal surface on which it slides?
Answer: 0.34

Problem 28: Rolling Ball on Incline

A solid ball of mass of 0.25 kg rolls without slipping 4 m down an incline that makes an angle of 25^0 a with a horizontal. If it starts from rest, what is its kinetic energy at the bottom of the incline?
A) 1 J B) 2 J, C) 4 J, D) 6 J, E) 8 J

Problem 29: Rolling Basketball

A basketball is rolling without slipping along a horizontal surface with total kinetic energy 20 J. How much energy (in Joules) is due to the rotational kinetic energy of the ball about its center of mass? Basketballs are hollow shells. (I = 2/3 mr^2.)
A) 10, B) 8, C) 13.3, D) 4, E) 12