Download End of Course Exam Review

End of Course Exam Review- ANSWER KEY
1.a. The object is moving in the positive direction only when there is a positive slope on a Position (displacement)
versus time graph. Therefore the object is moving in the positive (+) direction between 0-5 seconds.
1.b. With a slope of zero (Remember: slope on a position vs. time graph is the velocity of the object) the object is at
rest.
1.c. Velocity of the object is determined by calculating the slope of the line. Therefore: 𝑣 =
∆𝑥
∆𝑡
=
5−0
5−0
=1
𝑚
𝑠
**CORRECTION:In #2, all time units should read minutes based on provided graph.
2.a. On a velocity vs time graph the slope of the line represents the acceleration of the object. With a slope of zero, the
object is moving at a constant velocity in the positive (+) direction during this five minute interval. Also recall that
objects that are above the x-axis are in the positive direction and below the x-axis are in the negative direction.
2. b. Wherever there is a slope, the object is accelerating. Therefore: 0-10min, 15-30min, 30-40min, 40-55 min are all
intervals of acceleration.
2. c. Displacement and distance can both be determine on a velocity vs. time graph by analyzing the area underneath
the motion. To determine distance the SUM of the area underneath both above and below the x-axis would need to be
found. To determine displacement such as it is asking in this question, the DIFFERENCE in the areas underneath the
motion must be calculated. Therefore the total displacement of this object between 15-40 minutes is:
Above x-axis:
1
( × 15 × 60) = 450𝑚
2
Below x-axis:
1
× 10 × 40 = 200𝑚
2
∆𝑥 = 450 − 200 = +250𝑚
3.a. With a slope, the object is moving with a changing acceleration. Remember, acceleration has the ability to be in the
same direction or the opposite direction of velocity. Therefore, more information would be needed; such as a
corresponding velocity vs. time graph to truly determine direction. Possibilities include the object speeding up in the
positive direction or slowing down in the negative direction.
3.b. With no slope the object is moving with constant acceleration of +2 cm/s2. Again more information is needed to
determine whether it is speeding up in the positive direction or slowing down in the negative direction.
4. Distance and speed are scalar quantities. These quantities lack direction to them. Distance is the total ground
covered (regardless of direction) whereas the displacement of an object is its change in position (final position – initial
position). These quantities have the ability to be the same magnitude in basic linear motion but can also be different
when there is a change in direction.
Speed is just how “fast” an object is moving regardless of direction. Speed is based on the distance traveled per the
time of travel. Velocity is the rate at which on object changes position (displacement) per a given amount of time. This
value is a vector quantity where it includes direction. Again, velocity and speed can be the same magnitude in simple
motion scenarios but can also be different in more complex movements.
5.a. ∆𝑥 = 4300 𝑁 − 2200 𝑆 = 2100𝑚; 𝑁𝑜𝑟𝑡ℎ
5.b. 𝑑 = 4300 + 2200 = 6500 𝑚
CORRECTION: The man traveled for 1.5 hrs (this was missing from the question and is needed for part C and D)
5c. 𝑣 =
∆𝑥
∆𝑡
5.d. 𝑠 =
=
2100
5400
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑖𝑚𝑒
= 0.39
=
6500
5400
𝑚
𝑁𝑜𝑟𝑡ℎ
𝑠
= 1.2 𝑚/𝑠
6. If the motion of the object is a basic linear movement with no direction change the speed and velocity can be the
same magnitude because the distance and displacement are the same. For example, a car travels 500 meters north and
stops in 50 seconds. The distance and displacement are both 500 meters and time is 50 seconds regardless. Therefore,
500/50 both result in 10 m/s.
7. 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡
0 = 4.42 + (−0.75)∆𝑡
∆𝑡 = 5.9 𝑠
8. 𝑣𝑓2 = 𝑣𝑖2 + 2𝑎∆𝑥 (49.5)2 = 02 + 2(−9.81)∆𝑥
9. 𝑣𝑓 = 𝑣𝑖 + 𝑎∆𝑡 0 = 7.5 + (−9.81)∆𝑡
∆𝑥 = −125 𝑚 𝑜𝑟 125 𝑚 𝑑𝑜𝑤𝑛
∆𝑡 = 0.76 𝑠
10. Horizontally, the object’s initial velocity is completely in the horizontal direction. When air resistance is ignored (as it
was in our calculations), the object will move with the same velocity it is kicked with the entire fall. With a constant
horizontal velocity, there is no horizontal acceleration. The object will have a horizontal displacement (x) that it will land
out from the bridge. This is dependent on how fast it is kicked and the time it takes to fall.
Vertically, the object has no initial velocity. As it begins to descend, the stone will reflect that of a free-falling object
where it is accelerating due to gravity (-9.81 m/s2). It will displace vertically as it drops from the bridge. This object is
speeding up as it approaches the ground where it is gaining vertical velocity. The higher it drops from the faster it will
be moving as it approaches the ground and he longer it will take to reach it.
11. As a projectile is launched at an angle, it’s initial velocity has both vertical and horizontal components. Each can be
determined by using basic trigonometry functions. Observe the vector diagram of the projectile’s motion below.
1.2 m/s
𝑣𝑖𝑦 = 1.2𝑠𝑖𝑛14° = 0.29 m/s
14o
𝑣𝑖𝑥 = 1.2𝑐𝑜𝑠14° = 1.2 m/s
12. 1st : An object at rest remains at rest and an object in motion remains in motion unless an unbalanced force acts
upon it. Also known as the Law of Inertia, “objects want to keep doing what they’re doing.”
2nd : The net force on an object is dependent on the acceleration and mass an object possesses. The larger the mass or
acceleration the large the net force. Direct relationship.
3rd : For every action there is an opposite but equal reaction.
13. If the distance between two objects is doubled, the gravitational force between the objects would be ¼ as great due
to the following expression. The distance “r” is squared therefore by doubling the distance will quadruple this value.
𝐹𝑔 = 𝐺
𝑚1 𝑚2
𝑟2
𝐹𝑁 = 45616.5 𝑁 (𝐹𝑁 = 𝐹𝑔 )
𝐹𝑛𝑒𝑡 = 𝑚𝑎
14.
𝐹𝑛𝑒𝑡 = (4650)(8.51) = 39571.5 𝑁
𝐹𝑓 = 36493.2N
(45616.5*0.80)
Fapp = ???
𝐹𝑛𝑒𝑡 = 𝐹𝑎𝑝𝑝 − 𝐹𝑓
39571.5 = 𝐹𝑎𝑝𝑝 − 36493.2
𝐹𝑎𝑝𝑝 = 7.61𝑥104 𝑁; 𝑓𝑜𝑟𝑤𝑎𝑟𝑑
Fg = 45616.5 N
(4650*9.81)
15. CORRECTION: calculate the weight not mass.
𝑊 = 𝐹𝑔 = 𝑚𝑔 = (85.1)(9.81) = 835 𝑁; 𝑑𝑜𝑤𝑛
16.a. A billiard ball would drop faster than the cotton ball due to less air resistance acting on it.
16.b. If air resistance is ignored, the billiard ball and the cotton ball would fall at the same rate. (Recall: A vacuum is a
chamber with no air resistance)
17.
𝑐 2 = 15.02 + 4.52 𝑐 = 16 𝑚
15.0 m E
θ
4.5 m S
C
FN
Fapp
18.
Fy
Fg
Fx
Ff
Fg
𝑡𝑎𝑛𝜃 =
4.5
= 𝜃 = 17° 𝑆𝑜𝑢𝑡ℎ 𝑜𝑓 𝐸𝑎𝑠𝑡
15.0
19. Mechanical energy can be defined as the sum of the kinetic and potential energy an object possesses.
20.
∆𝐾𝐸 = 𝑊𝑛𝑒𝑡
1
1
𝑚𝑣𝑓2 − 𝑚𝑣𝑖2 = 𝑊𝑛𝑒𝑡
2
2
1
1
(0.25)(0)2 − (0.25)(0.76)2 = 𝑊𝑛𝑒𝑡
2
2
−0.72 𝐽
21.
Position #1
Position #2
𝑃𝐸 = (50)(9.81)(4) = 1962𝐽
𝑃𝐸 = (50)(9.81)(3) = 1471.5𝐽
𝐾𝐸 = 0𝐽
𝐾𝐸 = 𝑀𝐸 − 𝑃𝐸 = 490.5𝐽
𝑀𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1962 𝐽
𝑀𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1962 𝐽
𝑣 = 0𝑚/𝑠
1
490.5 = (50)𝑣 2 = 4.43 𝑚/𝑠
2
Position #3
𝑃𝐸 = (50)(9.81)(0) = 0𝐽
Position #4
1062 = (50)(9.81)(ℎ) = 2.3 𝑚
𝐾𝐸 = 𝑃𝐸1 = 1962𝐽
1
𝐾𝐸 = (50)(6)2 = 900𝐽
2
𝑀𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1962 𝐽
𝑀𝐸 = 𝑃𝐸 + 𝐾𝐸 = 1962 𝐽
1962 =
1
(50)𝑣 2 = 8.86 𝑚/𝑠
2
𝑃𝐸 = 𝑀𝐸 − 𝐾𝐸 (1962 − 900) = 1062𝐽