college physics exam 3 homework set solutions (kinematics)

College PhysicsExam 3 Homework Set Solutions

(Kinematics)For physics students everywhere!

Wayne HackerCopyright ©Wayne Hacker 2009. All rights reserved.

Physics 121 exam 3 homework set solns Copyright ©Wayne Hacker 2010. All rights reserved.1

Answer key for exam 3 (kinematics)

1. a b○ c d e 21. a b c d○ e 41. a○ b c d e

2. a○ b c d e 22. a b c○ d e 42. a○ b c d e

3. a b○ c d e 23. a b○ c d e 43. a b○ c d e

4. a○ b c d e 24. a b c○ d e 44. a○ b c d e

5. a b c d○ e 25. a b○ c d e 45. a b c d○ e

6. a b c d○ e 26. a b c○ d e 46. a b○ c d e

7. a b○ c d e 27. a○ b c d e 47. a b○ c d e

8. a b c d○ e 28. a b c○ d e 48. a b○ c d e

9. a○ b c d e 29. a b c d○ e 49. a b○ c d e

10. a b○ c d e 30. a○ b c d e 50. a b c d○ e

11. a b○ c d e 31. a b c d○ e

12. a b○ c d e 32. a b c d○ e

13. a○ b c d e 33. a b c d○ e

14. a○ b c d e 34. a b c○ d e

15. a○ b c d○ e 35. a b c d○ e

16. a○ b c d e 36. a b c d○ e

17. a b c○ d e 37. a b c○ d e

18. a b○ c d e 38. a b c○ d e

19. a b○ c d e 39. a b c d○ e

20. a○ b c d e 40. a○ b c d e


One-dimensional kinematics

-

6

t

x

sA

sB s

C@@@sD s

E

sF��sG

sH

Figure 1: A car is being driven erratically along a straight stretch of highway. The graphshows its position x as a function of the time t.

Problem 1. For the interval between points D and E in figure 1, the car’s velocity is:

(a) positive *(b) negative (c) zero

Solution: Since x decreases between D and E, the velocity is negative.

Problem 2. For the interval between points D and E in figure 1, the car’s accelerationis:

*(a) positive (b) negative (c) zero

Solution: The velocity is negative between C and D, and positive between E and F.Thus positive acceleration occurs between D and E.

-

6

t

x

sA s

B@@@sC s

D

sE

Figure 2: A car is being driven erratically along a straight stretch of highway. The graphshows its position x as a function of the time t.


Problem 3. For the interval between points C and D in figure 2, the car’s velocity is:

(a) positive *(b) negative (c) zero

Solution: Since x decreases between C and D, v is negative.

Problem 4. For the interval between points C and D in figure 2, the car’s accelerationis:

*(a) positive (b) negative (c) zero

Solution: The velocity between B and D is negative; the velocity between D and E ispositive. During the interval from C to D, the velocity goes from negative at point C tozero at point D. Since the velocity increases, a is positive.

Problem 5. A train is facing in the positive direction. It is rolling backward, and itsspeed is increasing. Which of the following describes the train’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) negative velocity; positive acceleration*(d) negative velocity; negative acceleration(e) none of these

Solution: The car is moving backward, so the velocity is negative. It is speeding up inthe backward direction, so the acceleration is negative. Careful! “Negative acceleration”and “slowing down” are not the same thing. In this case, negative acceleration meansspeeding up in the negative direction.

Problem 6. A car is facing in the positive direction. It is rolling backward down a hill,and its speed is increasing. Which of the following describes the car’s situation?

(a) positive velocity; positive acceleration(b) positive velocity; negative acceleration(c) negative velocity; positive acceleration*(d) negative velocity; negative acceleration(e) none of these

Solution: The car is moving backward, so the velocity is negative. It is speeding up inthe backward direction, so the acceleration is negative. Careful! “Negative acceleration”and “slowing down” are not the same thing. In this case, negative acceleration meansspeeding up in the negative direction.

Problem 7. If the upward direction is positive, which situation involves negative velocityand negative acceleration?

(a) An elevator moving upward at a constant speed*(b) An elevator speeding up as it moves downward(c) An elevator speeding up as it moves upward(d) An elevator slowing down as it moves downward(e) none of these

Solution: In order for velocity to be negative, the elevator must be moving downward.In order for acceleration to be negative, it must be speeding up in the downward direction.


Problem 8. If the upward direction is positive, which situation involves negative velocityand positive acceleration?

(a) An elevator moving upward at a constant speed(b) An elevator speeding up as it moves downward(c) An elevator speeding up as it moves upward*(d) An elevator slowing down as it moves downward(e) none of these

Solution: In order for velocity to be negative, the elevator must be moving downward.In order for acceleration to be positive, it must be slowing down as it moves downward.

Fundamental formulas for motion in 1-D

Equation 1: v = v0 + at (use when: position not in problem statement)

Equation 2: v2 = v20 + 2a(x− x0) (use when: time not in problem statement)

Equation 3: x = x0 + v0t+1

2at2 (use when: want position as a fcn of time)

Problem 9. (Fundamental Kinetic Equation 1) A boat is initially moving northwardat 3 m/s. It accelerates northward at 0.5 m/s2 for 12 s. What is the boat’s final speed?Round your answer to the nearest m/s.

*(a) 9 m/s (b) 36 m/s(c) 42 m/s (d) 72 m/s(e) None of these

Solution: We know acceleration, time, and initial velocity; we want the final velocity.

v = v0 + at = 3 m/s + (0.5 m/s2)(12 s) = 9 m/s

Problem 10. (Fundamental Kinematic Equation 2) Beginning at rest, a trainaccelerates at 0.10 m/s2 until it has gone 200 m. At the end of this time, how fast is thetrain moving? Round your answer to the nearest 0.1 m/s.

(a) 5.7 m/s *(b) 6.3 m/s(c) 7.0 m/s (d) 7.7 m/s(e) None of these

Solution: We know the initial position and initial velocity (both zero), the acceleration,and the final position. We want to know the final velocity. We have a formula that weuse when time isn’t in the problem statement.

v2 = v20 + 2a(x− x0) = 2ax

⇒ v =√

2ax =[2(0.10 m/s2)(200 m)

]1/2= 6.3 m/s


Problem 11. (Fundamental Kinetic Equation 3) A train is initially rolling south-ward at 3 m/s. It accelerates northward at 0.5 m/s2 for 20 s. At the end of this time,where is it relative to its starting point? Round your answer to the nearest meter.

(a) 80 m south *(b) 40 m north(c) 120 m north (d) 180 m north(e) None of these

Solution: We will take northward to be the positive direction, so the initial velocitywill be negative and the acceleration will be positive. We know the initial position (zero),the initial velocity, the acceleration, and the time; we want to know the final position.

x = x0 + v0t+1

2at2 = (−3 m/s)(20 s) +

1

2(0.5 m/s2)(20 s)2 = 40 m

Since the value of x is positive, it is north of the starting point.

Problem 12. While walking in the desert, you discover an old mine. To find out howdeep it is, you drop a rock down the shaft. Using the stopwatch on your cell phone,you time how long it takes before you hear the rock hit the bottom. If it takes 2.5 s forthe rock to reach the bottom, how deep is the shaft? Round your answer to the nearestmeter. For ease of calculation, assume that g = 10 m/s2.

(a) 21 m *(b) 31 m(c) 44 m (d) 63 m(e) None of these

Solution: We will take the positive direction to be downward, so that the accelerationand the final position will be positive. We will take the initial position to be zero. Thenwe know the initial position, the initial velocity (zero), the acceleration, and the time.We want to know the final position.

y = y0 + v0t+1

2at2 =

1

2at2 =

1

2(10 m/s2)(2.5 s)2 = 31 m

Problem 13. An astronaut hurls a watermelon straight downward from the top of ahigh cliff on Planet X, where the surface gravity is 3 m/s2. The initial speed of thewatermelon is 12 m/s downward. After 6 s, how fast is the watermelon moving?


Solution: We will take the positive direction to be downward, so the acceleration andthe initial velocity will be positive. We know the initial velocity, the acceleration, andthe time; we want to know the final velocity.

v = v0 + at = 12 m/s + (3 m/s2)(6 s) = 30 m/s


Problem 14. A banker drops a sack of pennies from the top of a building 20 m tall. Howfast is the sack moving when it strikes the ground? Round your answer to the nearestm/s. For ease of calculation, assume that g = 10 m/s2.


Solution: We will take the top of the building to be y0 = 0, and downward as thepositive direction. We know the initial and final position, the initial velocity (zero), andthe acceleration. We want the final velocity. We have a formula to use when time is notpart of the problem statement.

v2 = v20 + 2a(y − y0) = 2ay

⇒ v =√

2ay =

√2(10 m/s2)(20 m) = 20 m/s

Problem 15. A train is initially rolling forward at 15.0 m/s. It accelerates forward at0.5 m/s2. How long must it accelerate before it is moving forward at 20.0 m/s? Roundyour answer to the nearest second.

*(a) 10 s (b) 17 s(c) 20 s (d) 40 s(e) None of these

Solution: We know the initial velocity v0, the acceleration a, and the final velocity v.We want the time t. Use the first fundamental equation:

v = v0 + at ⇒ t =v − v0

a=

20.0 m/s− 15.0 m/s

0.5 m/s2 = 10 s

Problem 16. A train is moving at 24 m/s when the engineer is informed that there is apuppy on the track in front of him. If the train is capable of decelerating at 0.70 m/s2,how far does it go before it comes to a stop? Round your answer to the nearest 10 m.

*(a) 410 m (b) 450 m(c) 500 m (d) 550 m(e) None of these

Solution: We will take the forward direction to be positive, so the acceleration will benegative; and we will take the train’s initial position as x0 = 0. We know the initialposition, the initial velocity, the final velocity (zero), and the acceleration; we want toknow the final position. We have a formula to use when time is not in the statement ofthe problem.

v2 = v20 + 2a(x− x0)

⇒ v20 = −2ax

⇒ x =v2

0

−2a=

(24 m/s)2

−2(−0.70 m/s2)= 410 m


Problem 17. A train is initially going northward at 7.5 m/s. It accelerates at 0.3 m/s2

northward for 20 s. At the end of this time, how far north has it travelled? Round youranswer to the nearest 10 m.

(a) 120 m (b) 160 m*(c) 210 m (d) 260 m(e) None of these

Solution: We know the initial position (zero), the initial velocity, the acceleration, andthe time. We want to know the final position.

x = x0 + v0t+1

2at2 = (7.5 m/s)(20 s) +

1

2(0.3 m/s2)(20 s)2 = 210 m

Problem 18. A physics student on Planet X drops a watermelon from a cliff 35 m high.The watermelon hits the ground after 3.1 s. What is the surface gravity of Planet X?Round your answer to the nearest 0.1 m/s2.

(a) 6.6 m/s2 *(b) 7.3 m/s2

(c) 8.0 m/s2 (d) 8.8 m/s2

(e) None of these

Solution: We will take the positive direction to be downward. We will take the top ofthe cliff to be y0 = 0. We know the initial position and velocity (both zero), the finalposition, and the time. We want to know the acceleration.

y = y0 + v0t+1

2at2 =

1

2at2

⇒ a =2y

t2=

2(35 m)

(3.1 s)2= 7.3 m/s2


Problem 19. (Similarity Problem) An object dropped from a building with heighty1 takes time t1 to reach the ground. How long would it take for an object dropped froma building with height y2 = 4y1?

(a)√

2 t1 *(b) 2t1(c) 2

√2 t1 (d) 4t1

(e) None of these

Solution: We will take downward as the positive direction, and the top of the buildingas y0 = 0. The initial velocity is zero v0 = 0. Notice that we have two similar situationswith the fundamental equation (3) applying to both. Applying the initial conditions,equation (3) reduces to y = 1

2gt2. The common parameters to both situations are g and

2. Re-writing the governing equation with the common parameters on the right and thesituation-dependent variables on the left, gives

y =1

2gt2

· 2g−−−−−→ t2 =

2

gy

√−−−−−−→ t =

√2y

g√

y−−−−−→ t

√y

=

√2

g.

Substituting for y1, t1, and y2 yields

t2√y2

=

√2

g=

t1√y1

(where

√2

gis our common link)

· √y2−−−−−−→ t2 = t1

√y2

y1

= t1

√4y1

y1

= 2t1

Problem 20. (Similarity Problem) Starting from rest, a car rolls down a slope withconstant acceleration. After it has gone a distance x1, its speed is v1. What is its speedafter it has gone a distance of x2 = 2x1?

*(a)√

2 v1 (b) 2v1

(c) 2√

2 v1 (d) 4v1

(e) None of these

Solution: Since we’re not concerned with time in this problem, we’ll use the secondfundamental kinematic equation. Since v0 = 0 and x0 = 0, this reduces to v2 = 2ax. Wewill use downslope as the positive direction, so that all quantities will be positive. In thefirst situation, v2

1 = 2ax1. In the second,

v22 = 2a(2x1) = 2(2ax1) = 2v2

1 ⇒ v2 =√

2v21 =√

2 v1


One-dimensional rotational kinematics

Problem 21. Two bugs are clinging to the blade of a ceiling fan. The first bug is 10 cmfrom the center of the fan; the second bug is 20 cm from the center. If the first bug ismoving at a speed of v1, what is the speed of the second bug?

(a) v1/2 (b) v1

(c)√

2 v1 *(d) 2v1

(e) None of these

Solution: Use v = ωr. Both bugs are moving at the same angular speed ω, which isthe angular speed of the fan. If v1 = ω · (10 cm), then v2 = ω · (20 cm) = 2v1.

Problem 22. A windmill is initially held at rest by a brake. When the brake is released,it is subjected to an angular acceleration of 3 rad/s2. What angle will it turn throughbefore it reaches an angular speed of 15 rad/s?

(a) 5 rad (b) 18 rad*(c) 37.5 rad (d) 75 rad(e) None of these

Solution: We know the initial angular velocity ω0 = 0, the initial angle θ0 = 0, thefinal angular velocity, and the angular acceleration. We want to know the final angle.We neither know nor care about the time. We have a formula that we can use in thissituation:

ω2 = ω20 + 2α(θ − θ0) = 2αθ

⇒ θ =ω2

2α=

(15 rad/s)2

2(3 rad/s2)= 37.5 rad

Problem 23. A wind turbine is turning at a constant angular speed of 1.3 rad/s. Howlong does it take for the turbine to make one complete revolution? Round your answerto two significant figures.

(a) 4.1 s *(b) 4.8 s(c) 8.2 s (d) 9.7 s(e) None of these

Solution: One revolution = 2π rad; so

θ = ωt ⇒ =θ

ω=

2π rad

1.3 rad/s= 4.8 s


Problem 24. A fan is initially turning at 12 rad/s. The power to the fan is increased,applying a rotational acceleration of 2 rad/s2. What angle does the fan turn throughover the next 3 sec?

(a) 36 rad (b) 42 rad*(c) 45 rad (d) 126 rad(e) None of these

Solution: We know the initial angular speed ω0, the initial angle θ0 = 0, the angularacceleration α, and the time t. We want to know the final position θ. Use the thirdfundamental kinetic equation:

θ = θ0 + ω0t+1

2αt2 = (12 rad/s)(3 s) +

1

2(2 rad/s2)(3 s)2 = 45 rad

Problem 25. A shaft is initially turning at 10 rad/s. A brake is applied, producing anangular acceleration of −0.5 rad/s2. What angle does the shaft turn through before ithas slowed down to 8 rad/s?

(a) 32 rad *(b) 36 rad(c) 45 rad (d) 48 rad(e) None of these

Solution: We know ω0, ω, α, and θ0 = 0. We want to know θ. We don’t know and wedon’t care about t. Hence we use the equation that doesn’t involve t:

ω2 = ω20 + 2α(θ − θ0) = ω2

0 + 2αθ

⇒ θ =ω2 − ω2

0

2α=

(8 rad/s)2 − (10 rad/s)2

2(−0.5 rad/s2)= 36 rad


Two-dimensional kinematics

Problem 26. A car is driving at a constant speed along a winding road, running fromleft to right. Which of the following diagrams correctly shows the car’s velocity vectorat two points along the road?

-

6

x

y (a)ss

-

�

-

6

x

y (b)ss

6

?

-

6

x

y *(c)ss

-

-

-

6

x

y (d)ss

?

?

(e) None of these

Solution: The velocity vector is tangent to the curve and points in the direction ofmotion.


Problem 27. A car is driving at a constant speed along a winding road, running fromleft to right. Which of the following diagrams correctly shows the car’s acceleration vectorat two points along the road?

-

6

x

y *(a)ss

?6

-

6

x

y (b)ss

-

�

-

6

x

y (c)ss

6

? -

6

x

y (d)ss

�

-

(e) None of these

Solution: Since the speed is constant, the acceleration vector must be perpendicular tothe direction of motion: so (a) or (c). It must point inward (toward the concave side) oncurves; so (a).

Problem 28. A car is driving at a constant speed in aclockwise direction around a circular track. Which of thelabelled vectors is the velocity vector at point P?

(a) ~A (b) ~B

*(c) ~C (d) ~D(e) None of these

rP - ~B

6

~A

�~D

?~C

Solution: The velocity vector is tangent to the curve and points in the direction ofmotion.

Problem 29. A car is driving at a constant speed in aclockwise direction around a circular track. Which of thelabelled vectors is the acceleration vector at point P?

(a) ~A (b) ~B

(c) ~C *(d) ~D(e) None of these

rP - ~B

6

~A

�~D

?~C

Solution: In uniform circular motion, the acceleration vector points toward the centerof the circle.


Problem 30. The figure at right shows the tra-jectory of a cannonball. Which of the labelledvectors is the velocity vector at point P?

*(a) ~A (b) ~B

(c) ~C (d) ~D(e) None of these -

6

x

y rP ��3

~A

- ~BJJJ~C

?~D

Solution: Only ~A is tangent to the trajectory curve.

Problem 31. The figure at right shows the tra-jectory of a cannonball. Which of the labelledvectors is the acceleration vector at point P?

(a) ~A (b) ~B

(c) ~C *(d) ~D(e) None of these -

6

x

y rP ��3

~A

- ~BJJJ~C

?~D

Solution: The only acceleration in this situation is that of gravity, which is straightdownward. There is no horizontal component of acceleration.

Problem 32. The following four diagrams show the trajectory of a cannonball. Whichdiagram correctly shows the acceleration vectors at three points along the trajectory?

-

6

x

y(a)

rr r

6 ~0?

-

6

x

y(b)

rr r

��7

-

QQs

-

6

x

y(c)

rr r

ZZ~

? �

-

6

x

y*(d)

rr r

?

??

(e) None of these

Solution: Only figure (d) has acceleration vectors that point straight downward.


Uniform circular motion

Problem 33. A highway has a curve with a radius of 70 m. You attempt to drivearound the curve at 22 m/s. What is the centripetal acceleration on your car? Roundyour answer to the nearest 0.1 m/s2.

(a) 5.0 m/s2 (b) 5.6 m/s2

(c) 6.2 m/s2 *(d) 6.9 m/s2

(e) None of these

Solution: a =v2

r=

(22 m/s)2

70 m= 6.9 m/s2

Problem 34. A truck is going around a highway curve with a radius of 240 m. It willslide off the curve if it experiences a radial acceleration greater than 3.3 m/s2. How fastcan the truck safely go around the curve? Round your answer to the nearest m/s.

(a) 23 m/s (b) 25 m/s*(c) 28 m/s (d) 31 m/s(e) None of these

Solution: We know a and r; we want to know v.

a =v2

r⇒ v =

√ar =

√(3.3 m/s2)(240 m) = 28 m/s

Problem 35. (Similarity Solution) A train goes around a curve at speed v1, experi-encing a radial acceleration of a1. If a second train goes around the same curve at speedV2 = 2v1, what radial acceleration does it experience?

(a)√

2 a1 (b) 2a1

(c) 2√

2 a1 *(d) 4a1

(e) None of these

Solution: Let r1 be the radius of the curve (which doesn’t change). In the first situation,a1 = v2

1/r1. In the second situation,

a2 =(2v1)

2

r1=

4v21

r1= 4a1

Problem 36. A freeway on-ramp has a curve with a radius of 80 m. You attempt todrive around the curve at 15 m/s. What is the centripetal acceleration on your car?Round your answer to the nearest 0.1 m/s2.

(a) 2.4 m/s2 (b) 2.5 m/s2

(c) 2.7 m/s2 *(d) 2.8 m/s2

(e) None of these

Solution: a =v2

r=

(15 m/s)2

80 m= 2.8 m/s2


Problem 37. A curve in a highway has a radius of 90 m. A truck will roll over if itexperiences a centripetal acceleration greater than 2.5 m/s2. How fast can the truck drivearound the curve? Round your answer to the nearest m/s.

(a) 13 m/s (b) 14 m/s*(c) 15 m/s (d) 16 m/s(e) None of these

Solution: We know a and r; we want to know v.

a =v2

r⇒ v =

√ar =

√(2.5 m/s2)(90 m) = 15 m/s

Problem 38. (Similarity Problem) The Contention Railroad has an unbanked curvewith a radius of r1. Trains must go through the curve at a speed of v1 or less. If therailroad were rebuilt so that trains could negotiate it at v2 = 2v1, what would the curve’sradius r2 have to be?

(a)√

2 r1 (b) 2r1*(c) 4r1 (d) 16r1(e) None of these

Solution: Let v1 be the maximum speed of the train on a curve with radius r1. Thensince a = v2/r, the maximum centripetal acceleration that a train can safely experienceis a1 = v2

1/r1. Now we want to know how large r2 must be for a train going at a speedof v2 = 2v1 to experience an acceleration of a1. In this situation, the formula becomes

v21

r1= a1 =

(2v1)2

r2⇒ r2 =

(2v1)2

v22

r1 =4v2

1

v21

r1 = 4r1

Trajectories

half-parabola case

Problem 39. A movie stuntman wants to ride a motorcycle off a flat roof, sail acrossa gap 30 m wide, and land on another roof 4.0 m below the first one. How fast does heneed to be moving when he leaves the first roof? Round your answer to the nearest m/s.

(a) 28 m/s (b) 30 m/s(c) 32 m/s *(d) 33 m/s(e) None of these

Solution: The stuntman is going off a flat roof, with no mention of a ramp; so weassume that the angle of elevation is θ0 = 0. There is no vertical component of initialvelocity, and v0 = v0x.The time it takes for the motorcycle to fall y = 4.0 m is t =

√2y/g. During this time,

the motorcycle needs to cover a horizontal distance of x = 30 m. Hence the initial(horizontal) velocity must be

v =x

t=

x√2y/g

=

√g

2yx =

√9.8 m/s2

2(4.0 m)(30 m) = 33 m/s


Problem 40. A physics student wants to ride his bicycle off a flat roof, sail across a gap8.5 m wide, and land on another roof 2.5 m below the first one. How fast does he needto be moving when he leaves the first roof? Use g = 9.8 m/s2. Round your answer tothe nearest m/s.


Solution: The student is going off a flat roof, with no mention of a ramp; so we assumethat the elevation is θ0 = 0. There is no vertical component of initial velocity, andv0 = v0x.The time it takes for the bicycle to fall y = 2.5 m is t =

√2y/g. During this time, the

bicycle needs to cover a horizontal distance of x = 8.5 m. Hence the initial (horizontal)velocity must be

v =x

t=

x√2y/g

=

√g

2yx =

√9.8 m/s2

2(2.5 m)(8.5 m) = 12 m/s

Problem 41. A horizontal dock is 2.4 m above the water. A dog runs down the dockat 9.0 m/s and jumps off the end. How far from the end of the dock is the dog when itlands in the water? Use g = 9.8 m/s2. Round your answer to the nearest 0.1 m.

*(a) 6.3 m (b) 6.9 m(c) 7.6 m (d) 8.4 m(e) None of these

Solution: Since the dog is running off a horizontal dock, we assume that the elevation isθ0 = 0. Then the vertical component of the initial velocity is v0y = 0, and the horizontalcomponent is v0x = v = 13.4 m/s. The horizontal velocity does not change, since there’sno horizontal acceleration. Hence the horizontal distance that the dog travels beforehitting the water is the initial horizontal velocity times the time it takes to fall from thedock to the water:

x = v0xt = v0x

√2y

g= (9.0 m/s)

[2(2.4 m)

9.8 m/s2

]1/2

= 6.3 m


Problem 42. A bicyclist rides off a flat roof at 13.4 m/s. The roof is 3.6 m above theground. How far from the edge of the building does the bicyclist land? Round youranswer to the nearest meter.

*(a) 11 m (b) 12 m(c) 14 m (d) 15 m(e) None of these

Solution: Since the bicyclist is riding off a flat roof, we assume that the elevation isθ0 = 0. Then the vertical component of the initial velocity is v0y = 0, and the horizontalcomponent is v0x = v = 13.4 m/s. The horizontal velocity does not change, since there’sno horizontal acceleration. Hence the horizontal distance that the bicycle travels is theinitial horizontal velocity times the time it takes for the bicycle to fall to the ground:

x = v0xt = v0x

√2y

g= (13.4 m/s)

[2(3.6 m)

9.8 m/s2

]1/2

= 11 m

full-parabola case

Problem 43. You have designed a tennis-ball cannon with a muzzle velocity of 50 m/s.You want to land a tennis ball in a swimming pool 200 m away. At what angle to thehorizontal do you need to aim the gun? Round your answer to the nearest degree.

(a) 24◦ *(b) 26◦

(c) 28◦ (d) 30◦

(e) None of these

Solution: We know the initial velocity, the gravitational acceleration, and the range.We want the angle of elevation. Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ sin(2θ0) =

gxmax

v20

⇒ 2θ0 = sin−1

[gxmax

v20

]⇒ θ0 =

1

2sin−1

[gxmax

v20

]=

1

2sin−1

[(9.8 m/s2)(200 m)

(50 m/s)2

]= 26◦

Problem 44. You fire your potato gun from ground level at an angle of 30◦ above thehorizontal. The potato reaches a maximum height of 5.0 m. What was the initial speedof the potato? Round your answer to the nearest m/s.


Solution: We know the elevation, the gravitational acceleration, and the maximumheight; we want to know the initial velocity. Use the formula for maximum height:

ymax =1

2

v20

gsin2 θ0 ⇒ v0 =

√2gymax

sin2 θ0

=

√2(9.8 m/s2)(5.0 m)

sin2 30◦= 20 m/s


Problem 45. (Similarity Solution) You fire a cannon from ground level at a fixedelevation. The ball emerges from the muzzle with a speed of V ; it strikes the ground ata distance X from the cannon. You then increase the powder charge so that the muzzlevelocity of the ball is 2V . What is the new range of the ball?

(a)√

2X (b) 2X

(c) 2√

2X *(d) 4X(e) None of these

Solution: Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ X =

V 2

gsin(2θ0)

If we increase v0 from V to 2V , we get

xmax =(2V )2

gsin(2θ0) =

4V 2

gsin(2θ0) = 4X

Problem 46. Your potato gun has a muzzle velocity of 21 m/s. If you fire it at anelevation of 25◦, what is the maximum height reached by the potato? Use g = 9.8 m/s2.Round your answer to the nearest 0.1 m.

(a) 3.6 m *(b) 4.0 m(c) 4.5 m (d) 5.0 m(e) None of these

Solution: We have the initial velocity, elevation, and gravitational acceleration; wewant to know the maximum height. We have a formula for this:

ymax =1

2

v20

gsin2 θ0 =

(21 m/s)2(sin 25◦)2

2(9.8 m/s2)= 4.0 m

Problem 47. (Similarity Solution) You fire a cannon from ground level at an angleof elevation θ. The ball emerges from the muzzle with a speed of V , and strikes theground after time T . You then increase the powder charge in the cannon so that the ballemerges at a speed of 2V . How long does the ball remain in the air?

(a)√

2T *(b) 2T

(c) 2√

2T (d) 4T(e) None of these

Solution: We use the formula for the flight time: txmax =2v0 sin θ0

gThe flight time txmax and the initial velocity v0 change from the first situation to thesecond; the angle θ0 and the gravitational acceleration g do not. In the first situation,T = 2V sin θ0/g. In the second,

t2 =2(2V ) sin θ0

g= 2 · 2V sin θ0

g= 2T


Problem 48. A motorcyclist would like to ride up a short ramp on a rooftop, fly acrossa gap between two buildings, and land on a second rooftop at the same height as the first.The ramp is inclined at 32◦; the buildings are 18 m apart. How fast does the motorcyclisthave to be going when he makes the jump? Round your answer to the nearest m/s.

(a) 13 m/s *(b) 14 m/s(c) 15 m/s (d) 16 m/s(e) None of these

Solution: We know the gravitational acceleration, the elevation, and the range; wewant to know the initial velocity. Use the formula for range:

xmax =v2

0

gsin(2θ0) ⇒ v0 =

√gxmax

sin(2θ0)=

[(9.8 m/s2)(18 m)

sin(2 · 32◦)

]1/2

= 14 m/s

Problem 49. Your potato gun has a muzzle velocity of 23 m/s. You want to shoot apotato and have it strike the ground 3.5 seconds later. At what elevation do you have tofire the gun? Round your answer to the nearest degree.

(a) 43◦ *(b) 48◦

(c) 53◦ (d) 58◦

(e) None of these

Solution: We know the initial velocity, the gravitational acceleration, and the time offlight. We want to know the angle of elevation. Use the formula for flight time:

txmax =2v0 sin θ0

g⇒ sin θ0 =

gtxmax

2v0

⇒ θ0 = sin−1

[gtxmax

2v0

]= sin−1

[(9.8 m/s2)(3.5 s)

2(23 m/s)

]= 48◦


Problem 50. (Lab Similarity Problem) You fire a spring cannon from ground levelat a fixed elevation. The spring cannon uses a compressed spring to launch the ball.The cannon has three settings with the first being the lowest and the third being thestrongest. Using the first setting (one click) you find that the ball emerges from themuzzle with a speed of V ; it strikes the ground at a distance X from the cannon. Youthen increase the energy stored in the spring by using the second setting (two clicks).The muzzle velocity of the ball is now αV , where α is a positive constant. What is thenew range of the ball?

(a)√αX (b) αX

(c) α√αX *(d) α2X

(e) None of these

Solution: Use the formula (the governing equation) for range:

xmax =v2

0

gsin(2θ0) ⇒ xmax,1 =

V 2

gsin(2θ0)

If we increase v0 from V to αV , we get

xmax,2 =(αV )2

gsin(2θ0) = α2

(V 2

gsin(2θ0)

)= α2xmax,1 = α2X

Thus, if we double the initial speed we quadruple the distance traveled by the ball.Actually, in the real world the faster the ball travels, the larger the air drag.

college physics exam 3 homework set solutions (kinematics)

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