Physics - OAK PARK USD
AP Physics 2: Kinematics—Two Dimension Name __________________________
A. Vectors
1. vectors and scalars
a. d, v and a have magnitude (how much) and direction (which way) ∴ vectors
b. t, m, distance, speed only have magnitude ∴ scalars
c. arrow is used to symbolize a vector
1. arrow points in the vector’s direction
2. magnitude ∝ arrow's length
2. addition of vectors—tail to tip method
vectors are laid out tail to tip; the sum (resultant) equals the length and angle of the line that connects the tail of the first vector to the tip of the last vector
B
R = A + B
A
3. addition of vectors—component method
|an x-y coordinate system is established and θ is measured |
|counterclockwise from +x-axis (0o) |
|y = 90o |
| |
| |
|B |
|θB |
| |
|A |
|θA |
|x = 0o |
|x-component and y-component for each vector are calculated (R = |
|magnitude; θ = direction) |
|cosθ = adjacent/hypotenuse ∴ Rx = Rcosθ |
|sinθ = opposite/hypotenuse ∴ Ry = Rsinθ |
|y Rx = Ax + Bx Bx= BcosθB |
| |
|B By = BsinθB |
| |
|Ry = Ay + By R |
| |
|Ay = AsinθA |
|A |
|x |
|Ax = AcosθA |
|Rx = Ax + Bx = AcosθA + BcosθB |
|Ry = Ay + By = AsinθA + BsinθB |
|R = (Rx2 + Ry2)½ |
|tanθ = Ry/Rx ∴ θ = tan-1(Ry/Rx) |
|add 180o to θ when Rx is negative |
4. relative motion
a. observed velocity of an object depends on the motion of the observer relative to the object
b. observed acceleration is the same for any non-accelerating observer
c. example: a boat is traveling vboat with respect to the water, but the water current is vwater with respect to the Earth ∴ the boat velocity with respect to the Earth is vector sum: v = vboat + vwater
river current → vboat vboat + vwater
vwater
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B. Projectile Motion
1. projectile's horizontal velocity is constant
2. projectile's vertical velocity is immediately affected by the downward acceleration of gravity
3. solving projectile motion problems
a. general solution
|determine vxo = vocosθ and vyo = vosinθ |
|complete the "y" row in the data chart for all values in the vertical |
|direction (↓ is negative) |
|complete the "x" row in the data chart for all values in the horizontal |
|direction (vxo = vx) |
|time is the same for y and x directions |
|direction |
|d |
|vo |
|vt |
|a |
|t |
| |
|y |
|dy |
|vyo |
|vyt |
|-10 m/s2 |
|t |
| |
|x |
|dx |
|vx |
| |
| |
| |
|solve for unknown in the vertical direction with |
|dy = vyot + ½gt2 |
|vyt = vyo + gt |
|vyt2 = vyo2 + 2gdy |
| |
|no vyt |
|no dy |
|no t |
| |
|solve for unknown in the x direction with dx = vxt |
b. projectile launched horizontally from height h
|data chart |
|direction |
|d |
|vo |
|vt |
|a |
|t |
| |
|y |
|-h |
|0 |
|vyt |
|-10 m/s2 |
|t |
| |
|x |
|dx |
|vx = vo |
| |
| |
| |
|solve for unknown in the vertical direction with |
|h = ½gt2 |
|vyt = gt |
|vyt2 = 2gh |
| |
|solve for unknown in the x direction with dx = vxt |
c. projectile launched at angle θo from height 0
|determine vxo = vocosθo and vyo = vosinθo |
|data chart—highest point |
|direction |
|d |
|vo |
|vt |
|a |
|t |
| |
|y |
|dy |
|vyo |
|0 |
|-10 m/s2 |
|t |
| |
|x |
|dx |
|vx |
| |
| |
| |
|solve for unknown in the vertical direction with |
|vyo = gt |
|0 = vyo2 + 2gdy |
| |
|no dy |
|no t |
| |
|solve for unknown in the x direction with dx = vxt |
|data chart—landing point |
|direction |
|d |
|vo |
|vt |
|a |
|t |
| |
|y |
|0 |
|vyo |
|-vyo |
|-10 m/s2 |
|t |
| |
|x |
|dx |
|vx |
| |
| |
| |
|solve for unknown in the vertical direction with |
|2vyo = gt |
| |
|solve for unknown in the x direction with dx = vxt |
d. projectile launched at angle θo from height h
|determine vxo = vocosθ and vyo = vosinθ |
|data chart—landing point |
|direction |
|d |
|vo |
|vt |
|a |
|t |
| |
|y |
|dy |
|vyo |
|vyt |
|-10 m/s2 |
|t |
| |
|x |
|dx |
|vx |
| |
| |
| |
|solve for unknown in the vertical direction with |
|dy = vyot + ½at2 |
|vyt = vyo + at |
|vyt2 = vyo2 + 2ady |
| |
|no vyt |
|no dy |
|no t |
| |
|solve for unknown in the x direction with dx = vxt |
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A. Vectors
Questions 1-12 Briefly explain your answer.
1. If two vectors A + B = 0, what can you say about the magnitude and direction of the two vectors?
(A) perpendicular
(B) parallel in the opposite direction
(C) parallel in the same direction
|B—with only two vectors, they must exactly cancel out ∴ up and back |
2. Given vectors A + B = C and |A|2 + |B|2 = |C|2, how are the vectors A and B oriented with respect to each other?
(A) perpendicular
(B) parallel in the opposite direction
(C) parallel in the same direction
|A—second equation is true only for 45-45-90 triangle ∴ vectors must be |
|perpendicular |
3. Given vectors A + B = C and |A| + |B| = |C|, how are the vectors A and B oriented with respect to each other?
(A) perpendicular
(B) parallel in the opposite direction
(C) parallel in the same direction
|C—The only way the negative and positive case don't matter is when the |
|two vectors are in the same direction |
4. If each component of a vector is doubled, what happens to the angle of the vector?
(A) doubles (B) the same (C) reduced by half
|B—the two side are doubled, but the triangles are similar ∴ same angle |
|(tanθ = y/x = 2y/2x) |
5. If each component of a vector is equal in length, what is the angle of the vector?
(A) 30o (B) 45o (C) 60o (D) 30o or 60o
|B—Two sides of a right triangle are equal, then non-right angles are |
|also equal (tanθ = 1) |
6. If one of the components is half the length of the vector, what is the angle of the vector?
(A) 30o (B) 45o (C) 60o (D) 30o or 60o
|D—The sides for a 30-60-90 triangle are 1-2-√3, but you don't know which|
|side is half the length so 30 or 60 |
Questions 7-8 Vx = 30 units Vy = 40 units.
7. What is the magnitude of the resultant V?
(A) 30 units (B) 40 units (C) 50 units (D) 70 units
|C—This is a 3-4-5 triangle, with 5 being the hypotenuse |
8. What is the angle of the resultant V?
(A) 37o (B) 45o (C) 53o (D) 37o or 53o
|C—The angles for a 3-4-5 triangle are 37o-53o-90o, with the 53o opposite|
|the 4 side |
9. You are adding vectors of length 20 units and 40 units. What is the only possible resultant magnitude?
(A) 0 (B) 18 (C) 37 (D) 64
|C—The vector sum can be any value between 40-20 and 40+20, which is |
|between 20 and 60, which fits C |
Questions 10-11 Two students are tossing a ball back and forth in the aisle of a train. The velocity of the ball is vB and the velocity of the train is vT.
10. What is the velocity of the ball with respect the ground if the velocities of the ball and train are in the same direction?
(A) vB + vT (B) vB – vT (C) 0
|A—The ball has the train's velocity and its own in the same direction ∴ |
|add |
11. What is the velocity of the ball with respect the ground if the velocities of the ball and train are in the opposite direction?
(A) vB + vT (B) vB – vT (C) 0
|B—The ball has the train's velocity and its own, but they are in the |
|opposite direction ∴ subtract |
12. A student jogs 100 m at 60o north of east. What are the x- and y-components of the displacement vector?
|dx = (100 m)cos60 = 50 m |
|dy = (100 m)sin60 = 86.6 m |
13. Determine the resultant for each situation below.
| |magnitude |direction |
|6 m ↑ + 8 m ↑ |6 m + 8 m = 14 m |north (90o) |
|6 m ↑ + 8 m ↓ |6 m + (-8 m) = -2 m |south (-90o) |
|6 m ↑ + 8 m ← |(62 + -82)½ = 10 m |tanθ = |6/-8| = -37o |
| | |∴ 180 – 37 = 143o |
14. a. Label the x-y grid below:
(1) north, south, east, west
(2) 0o, 90o, 180o, and 270o
b. Draw a displacement vector of 20 m 30o north of east
(1 cm = 10 m scale)
15. Adding Vectors
a. Measure the magnitude and angle of vector A and B
B
A
|Vector |Magnitude (L) |Angle (θ) |
|A |3.1 cm |47.0o |
|B |3.5 cm |-26.5o |
b. Calculate the x- and y-components of the two vectors and then add the x-components to determine the resultant
x-component and repeat with the y-components.
|Vector |x-component |y-component |
|A |3.1cos(47.0) = 2.1 cm |3.1sin(47.0) = 2.3 cm |
|B |3.5cos(-26.5) = 3.1 cm |3.5sin(-26.5) = -1.6 cm |
|R |5.2 cm |0.7 cm |
c. Calculate the magnitude and angle of the resultant.
|Vector |magnitude |direction |
|R |(5.22 + 0.72)½ = 5.2 cm |θ = tan-1(0.7/5.2) = 8o |
d. Draw the resultant vector on the diagram and measure its magnitude and angle.
|Vector |Magnitude |Angle |
|R |5.3 cm |7o |
e. Determine the percent difference between the two.
|Magnitude |Angle |
|100(5.2 – 5.3)/5.2 = 2 % |100(8 – 7)/8 = 13 % |
16. Consider the following vectors, A and B.
|Vector |A |B |
|Magnitude |8.0 cm |4.0 cm |
|Angle |25o |150o |
a. Calculate the x-component and y-component of each vector and then determine the x-component and y-component of the resultant R.
|Vector |x-component |y-component |
|A |8.0cos(25) = 7.3 |8.0sin(25) = 3.4 |
|B |4.0cos(150) = -3.5 |4.0sin(150) = 2.0 |
|R |3.8 cm |5.4 cm |
b. Calculate the magnitude and angle of the resultant.
|Vector |Magnitude |Angle |
|R |(3.82 + 5.4)½ = 6.6 cm |tan-1(5.4/3.8) = 55o |
17. Consider the following vectors, A, B and C.
|Vector |Magnitude |Angle |
|A |12 cm |45o |
|B |8 cm |135o |
|C |10 cm |-75o |
a. Calculate the x- and y-component of vectors A, B and C and then determine the resultant's components.
|Vector |x-component |y-component |
|A |12cos(45) = 8.5 |12sin(45) = 8.5 |
|B |8cos(135) = -5.7 |8sin(135) = 5.7 |
|C |10cos(-75) = 2.6 |10sin(-75) = -9.7 |
|R |5.4 cm |4.5 cm |
b. Calculate the magnitude and angle of the resultant.
|Vector |magnitude |direction |
|R |(5.42 + 4.52)½ = 7.0 cm |tan-1(4.5/5.4) = 40o |
18. A boat travels at 1.85 m/s (vboat) in a river whose current (vwater) is 1.20 m/s west. Determine the velocity of the boat with respect to the ground for each situation.
a. the boat is headed due east?
|v = vboat + vwater = 1.85 m/s + (-1.20 m/s) = 0.65 m/s |
|east ∴ θ = 0o |
b. the boat is headed due north?
|v = (vboat2 + vwater2)½ = [(1.85)2 + (1.20)2]½ = 2.21 m/s |
|tanθ = 1.85/-1.20 = -57o ∴ 180 + -57 = 123o |
Questions 19-20 Alice and Bill wish to cross a river, which is 800 m wide with a current of 3 m/s south (vwater) in boats with a maximum speed of 4 m/s (vboat)
19. Alice directs her boat east but ends up going south east.
a. Label vboat, vwater and vboat + vwater.
vboat
vwater
vboat + vwater
b. What is the resultant velocity of the boat in the water?
|v = (vboat2 + vwater2)½ |
|v = [(4 m/s)2 + (3 m/s2)]½ = 5 m/s |
c. What is the direction of the boat with respect to Earth?
|tanθ = -3/4 |
|θ = -37o (south east) |
d. How much time did it take Alice to cross the river?
|d = vt |
|800 m = (4 m/s)t ∴ t = 200 s |
e. How far down stream did the boat travel?
|d = vt = (3 m/s)(200 s) = 600 m |
20. Bill directs his boat north east but ends up going east.
a. Label vboat, vwater and vboat + vwater.
vboat
vwater
vboat + vwater
b. What is the resultant velocity of the boat in the water?
|v = (vboat2 – vwater2)½ |
|v = [(4 m/s)2 – (3 m/s2)]½ = 2.6 m/s |
c. What is the direction that Bill heads the boat?
|sinθ = 3/4 |
|θ = 49o (north east) |
d. How much time did it take to cross the river?
|d = vt |
|800 m = (2.6 m/s)t ∴ t = 308 s |
21. A FedEx plane travels from New York to Los Angeles (4,000 km), stays in LA for 90 minutes to refuel and unload and reload cargo, and then returns to NY. The plane's air speed is 800 km/hr and the west to east wind current is 100 km/hr. How long does it take for the round trip?
|NY to LA: t = d/v = (4000 km)/(800 – 100)km/hr = 5.7 hr |
|LA to NY: t = d/v = (4000 km)/(800 + 100)km/hr = 4.4 hr |
|total time: 5.7 hr + 1.5 hr + 4.4 hr = 11.6 hrs |
B. Projectile Motion
Questions 22-35 Briefly explain your answer.
22. A small cart is rolling a constant velocity on a flat track. It fires a ball straight up into the air as it moves. In relation to the cart, the ball lands
(A) In front (B) in the cart (C) behind
|B—the launched ball has the same horizontal velocity as the cart, so it |
|is always directly above the cart |
23. A small cart accelerates on a flat track. It fires a ball straight up into the air as it speeds up. In relation to the cart, the ball lands
(A) In front (B) in the cart (C) behind
|C—the launched ball has the same horizontal velocity as the cart when it|
|was launched, but the cart speed up |
24. You drop a package from a plane flying at constant speed in a straight line. Without air resistance, the package will
(A) lag behind the plane as it falls
(B) remain directly beneath the plane as it falls
(C) move ahead of the plane as it falls
|B—the package and plane have the same horizontal velocity ∴ the package |
|keeps up with the plane |
25. A projectile is launched at an angle of 30o. At what point in its trajectory does the projectile have the lowest speed?
(A) at the start (B) at the middle (C) at the end
|B—it is slowest when it reaches the highest point, where vy = 0 (vx is |
|constant) |
Questions 26-27 From the same height and at the same time, ball A is dropped and ball B is fired horizontally.
26. Which ball will land first?
(A) A (B) B (C) tie
|C—time only depends on vertical drop, which is the same for both |
27. Which ball has the greater velocity when it lands?
(A) A (B) B (C) tie
|B—the overall velocity is vector sum of vx and vy, since B has greater |
|vx, its overall velocity is greater |
28. Which punt has the longest hang time, or do they tie?
A B C
|tie—since the balls reach the same height, they have equal hang time |
29. Which throw from the outfield will reach second base first?
A B second base
|A—it doesn't go as high ∴ it spends less time in the air |
30. The two identical projectiles use the same launcher. Path A takes place on Earth and path B on the Moon, where g is 1/6 Earth's.
A B
Which accounts for the greatest difference between the two paths?
(A) The initial velocity on the Moon is greater than on Earth.
(B) The downward acceleration on the Moon is less than on Earth.
(C) The airless environment on the Moon allows for the horizontal velocity to remain constant.
|B—with 1/8 the acceleration due to gravity, the ball's vy doesn't slow |
|down as quickly and the ball goes higher |
31. A carnival game includes a gun and a hanging target, which is released at the instant the gun is discharged. In order to strike the target you should aim
(A) above the target
(B) at the target
(C) below the target
|B—the bullet and target will descend with equal acceleration downward ∴ |
|aim at the target |
32. A ball is thrown horizontally at 10 m/s from a bridge that is 20 m above the water.
a. Complete the chart
| |d |vo |vt |a |t |
|y |-20 m |0 m/s |vyt |-10 m/s2 |t |
|x |dx |10 m/s | | |
b. How long does it take for the ball to hit the water?
|dy = vyot + ½at2 |
|20 m = ½(-10 m/s2)t2 ∴ t = 2 s |
c. How far horizontally does the ball travel?
|dx = vxot = (10 m/s)(2 s) = 20 m |
33. A rescue plane (traveling horizontally at 70 m/s) wants to drop supplies to climbers on a rocky ridge 235 m below.
a. Complete the chart
| |d |vo |vt |a |t |
|y |-235 m |0 m/s |vyt |-10 m/s2 |t |
|x |dx |70 m/s | | |
b. How long does it take for the supplies to fall 235 m?
|dy = vyot + ½at2 |
|235 m = ½(-10 m/s2)t2 ∴ t = 6.9 s |
c. How far in advance must the supplies be dropped?
|dx = vxot = (70 m/s)(6.9 s) = 483 m |
d. What is the line of sight angle from the plane to ridge?
|θ = tan-1(dy/dx)| = tan-1(-235/483) = -26o |
e. What is the vertical velocity, vyt?
|vyt = vyo + at = (-10 m/s2)(6.9 s) = -69 m/s |
f. With what speed do the supplies land?
|vt = (vxt2 + vyt2)½ = (702 + 692)½= 98 m/s |
34. A ball is thrown from the top of a 20 m building with an initial velocity of 10 m/s at various angles. Complete the charts.
a. The ball is thrown at 0o
| |d |vo |vt |a |t |
|y |-20 m |0 m/s |vyt |-10 m/s2 |t |
|x |dx |10 m/s | | |
|vyt|vy2 = vyo2 + 2ady = 0 + 2(-10)(-20) |
| |vy = -20 m/s |
|t |vy = vyo + at |
| |-20 = 0 + (-10)t ∴ t = 2 s |
|dx |dx = vxot |
| |dx = (10)(2) = 20 m |
|vt |v = (-202 + 102)½ |
| |v = 22 m/s |
b. The ball is thrown at 90o
| |d |vo |vt |a |t |
|y |-20 m |10 m/s |vyt |-10 m/s2 |t |
|x |dx |0 m/s | | |
|vyt|vy2 = vyo2 + 2ady = 102 + 2(-10)(-20) |
| |vy = -22 m/s |
|t |vy = vyo + at |
| |-20 = 10 + (-10)t ∴ t = 3 s |
c. The ball is thrown at -90o
| |d |vo |vt |a |t |
|y |-20 m |-10 m/s |vyt |-10 m/s2 |t |
|x |dx |0 m/s | | |
|vyt|vy2 = vyo2 + 2ady = (-10)2 + 2(-10)(-20) |
| |vy = -22 m/s |
|t |vy = vyo + at |
| |-20 = -10 + (-10)t ∴ t = 1 s |
d. The ball is thrown at 37o
| |d |vo |vt |a |t |
|y |-20 m |6 m/s |vyt |-10 m/s2 |t |
|x |dx |8 m/s | | |
|vyt|vy2 = vyo2 + 2ady = (6)2 + 2(-10)(-20) |
| |vy = -21 m/s |
|t |vy = vyo + at |
| |-20 = 6 + (-10)t ∴ t = 2.6 s |
|dx |dx = vxot |
| |dx = (8)(2.6) = 21 m |
|vt |v = (-212 + 82)½ |
| |v = 22 m/s |
35. A soccer ball is kicked 20 m/s at an angle of 30o.
a. Determine the x- and y- components of the velocity vo?
|vxo |vx0 = v0cosθ = 20cos30 = 17 m/s |
|vyo |vy0 = v0sinθ = 20 sin30 = 10 m/s |
b. Complete the chart for the ball when it lands.
| |d |vo |vt |a |t |
|y |0 m |10 m/s |-10 m/s |-10 m/s2 |t |
|x |dx |17 m/s | | |
c. What is the total time in the air?
|vy = vyo + at |
|-10 = 10 + (-10)t ∴ t = 2 s |
d. What is the total distance the ball travels horizontally?
|dx = vxot = (17)(2) = 34 m |
e. Complete the chart for the ball at its highest point.
| |d |vo |vt |a |t |
|y |dy |10 m/s |0 m/s |-10 m/s2 |1 s |
|x |
g. Graph the following.
| |Displacement |Velocity |Acceleration |
|Horizontal | |t | |t | |t |
| | | | | | | |
| | | | | | | |
|Vertical | |t | | |t | |t |
| | | | | | | | |
| | | | | | | | |
36. A projectile, fired with velocity vo and angle θ, lands on a cliff, which is 195 m away and 155 m high, in 7.6 s.
a. Complete the chart
| |d |vo |vt |a |t |
|y |155 m |vyo |vyt |-10 m/s2 |7.6 s |
|x |195 m |vxo | | |
b. What is vxo?
|vxo = dx/t = 195/7.6 = 26 m/s |
c. What is vyo?
|dy = vyot + ½at2 |
|155 = vyo(7.6) + ½(-10)(7.6)2 ∴ vy0 = 58 m/s |
d. What is vo?
|v02 = vxo2 + vyo2 = 262 + 582 = 64 m/s |
e. What is θ?
|tanθ = vyo/vxo = 58/26 = 2.2 ∴ θ = 66o |
Catapult Lab
37. Determine the optimum setting (listed below) for the ping-pong ball catapult by measuring the distance in the air.
L1: length of arm from rubber band to axel
L2: length of arm from cup to axel
θ: launch angle
A: rubber band anchor
a. Complete the data chart
|L1 |L2 |θ |A |Distance |
| | | | |1 |2 |3 |Avg |
|1 |4 |8 |11 | | | | |
|2 |4 |8 |11 | | | | |
|3 |4 |8 |11 | | | | |
|Max |5 |8 |11 | | | | |
| |6 |8 |11 | | | | |
| |7 |8 |
| | | | | | |
c. Calculate the following from the data.
| |Formula |Calculation |
|vxo |vxo = dx/t | |
|vyo |dy = vyot + ½at2 | |
|vo |vo = (vxo2 + vyo2)½ | |
|θ |tanθ = vyo/vxo | |
38. A ball is thrown horizontally at 20 m/s from a bridge that is 80 m above the water.
| |d |vo |vt |a |t |
|y |80 m |0 m/s |vyt |-10 m/s2 |t |
|x |dx |20 m/s | | |
a. How long does it take for the ball to hit the water?
|dx= vot + ½at2 |
|80 m = 0 + ½(-10 m/s2)t2 ∴ t = 4 s |
b. How far horizontally does the ball travel?
|dx = vt = (20 m/s)(4 s) = 80 m |
39. A rescue plane, traveling horizontally at 100 m/s, wants to drop supplies to climbers on a rocky ridge 125 m below.
| |d |vo |vt |a |t |
|y |-125 m |0 m/s |vyt |-10 m/s2 |t |
|x |dx |100 m/s | | |
a. How long does it take for the supplies to fall 125 m?
|dx = vot + ½at2 |
|125 m = 0 + ½(-10 m/s2)t2 ∴ t = 5 s |
b. How far in advance must the supplies be dropped?
|dx = vt = (100 m/s)(5 s) = 500 m |
c. What is the line of sight angle from the plane to ridge?
|θ = tan-1(dy/dx)| = tan-1(-125/500) = -14o |
d. What is the vertical velocity vyt?
|vy = vo + at = 0 + (-10 m/s2)(5 s) = -50 m/s |
e. With what speed do the supplies land?
|v = (vx2 + vy2)½ = (1002 + 502)½ = 112 m/s |
40. A soccer ball is kicked 30 m/s at an angle of 60o.
a. Determine the x- and y- components of the velocity vo.
|vxo |vxo = vocosθ = (30 m/s)cos60 = 15 m/s |
|vyo |vyo = vosinθ = (30 m/s)sin60 = 26 m/s |
b. Complete the chart for the ball when it lands.
| |d |vo |vt |a |t |
|y |0 m |26 m/s |-26 m/s |-10 m/s2 |t |
|x |dx |15 m/s | | |
c. What is the total time in the air?
|vy = vyo + at |
|-26 m/s = 26 m/s + (-10 m/s2)t ∴ t = 5.2 s |
d. What is the total distance the ball travels horizontally?
|dx = vxot = (15 m/s)(5.2 s) = 78 m |
e. What is the maximum height?
|vy2 = vyo2 + 2ady |
|02 = (26 m/s)2 + 2(-10 m/s2)dy ∴ dy = 34 m |
41. A student throws a ball that lands on a cliff 60 m away and 20 m above the student. The ball is in the air for 4 s.
| |d |vo |vt |a |t |
|y |20 m |vyo |vy |-10 m/s2 |4 s |
|x |60 m |vxo | | |
Determine the following for the ball's initial velocity
a. horizontal component
|vx = dx/t = 60 m/4 s = 15 m/s |
b. vertical component
|dy = vyot + ½at2 |
|20 m = vyo(4 s) + ½(-10 m/s2)(4 s)2 ∴ vyo = 25 m/s |
c. magnitude
|(vx2 + vy2)½ = (152 + 252)½ = 29 m/s |
d. direction
|tanθ = vy/vx = 25 m/s/15 m/s = 59o |
Practice Multiple Choice (no calculator)
Briefly explain why the answer is correct in the space provided.
|1 |
2. Two spheres, A and B, are thrown horizontally from the top of a tower; vA = 40 m/s and vB = 20 m/s. Which is true of the time T in the air and horizontal distance d traveled?
(A) TA = TB, dA = dB (B) TA = TB, dA = 2dB
(C) TA = TB, 2dA = dB (D) TA > TB, dA > dB
|time is the same because they fall the same distance. Horizontal |
|distance is greater for A because greater vx |
3. A plane flying horizontally at 100 m/s drops a crate. What is the horizontal component of the crate's velocity just before it strikes the ground 3 seconds later?
(A) 0 m/s (B) 100 m/s (C) 300 m/s (D) 400 m/s
|the crate and plane have same horizontal velocity (assuming that air |
|resistance is negligible) |
Questions 4-5 A ball is launched horizontally at 5 m/s from a height of 10 m.
5 m/s
10 m
4. How much time is the ball in the air?
(A) 0.5 s (B) 0.7 s (C) 1.4 s (D) 2 s
|d = vot + ½at2 |
|10 m = 0 + ½(-10 m/s2)t ∴ t = √2 = 1.4 s |
5. How far does the ball travel horizontally before hitting the ground?
(A) 2.5 m (B) 5 m (C) 7 m (D) 10 m
|dx = (vx)(t) = (5 m/s)(1.4 s) = 7 m |
6. A projectile is fired with initial velocity, vo, at an angle θo.
[pic]
Which pair of graphs represent the vertical components of the velocity vy and acceleration ay of the projectile as functions of time t?
(A) vy ay (B) vy ay
t t t t
(C) vy ay (D) vy ay
t t t t
|vertical velocity goes from +vo to – vo and acceleration is constant -10|
|m/s2 during the entire flight. |
Questions 7-10 A projectile is fired from the ground with an initial velocity of 250 m/s at an angle of 37o above horizontal.
7. What is the vertical component of the initial velocity?
(A) 100 m/s (B) 150 m/s (C) 200 m/s (D) 250 m/s
|given the angle of 37o, then this is a 3-4-5 triangle, |
|∴ vy = 3/5(250 m/s) = 150 m/s. |
8. How long is the projectile in the air (assume the projectile lands at the same elevation that it was fired)?
(A) 25 s (B) 30 s (C) 45 s (D) 50 s
|vertical velocity goes from +vo to -vo while in the air. |
|vt = vo + at → -150 m/s = 150 m/s + (-10 m/s2)t ∴ t = 30 s |
9. What is the horizontal component of the initial velocity?
(A) 100 m/s (B) 150 m/s (C) 200 m/s (D) 250 m/s
|vx = 4/5( 250 m/s) = 200 m/s |
10. How far did the projectile travel horizontally before it struck the ground?
(A) 6,000 m (B) 7,000 m (C) 9,000 m (D) 10,000 m
|dx = (vx)(t) = (200 m/s)(30 s) = 6,000 m |
11. A projectile is fired with an initial velocity of 100 m/s at an angle θ above the horizontal. If the projectile's initial horizontal speed is 60 m/s, then angle θ measures
(A) 30o (B) 37o (C) 40o (D) 53o
|this makes a 3-4-5 triangles, ∴ θ = sin-1(60/100) = 53o |
12. A rock is dropped from the top of a 45-m tower, and at the same time a ball is thrown from the top of the tower in a horizontal direction. The ball and the rock hit the level ground a distance of 30 m apart. The horizontal velocity of the ball thrown was most nearly
(A) 5 m/s (B) 10 m/s (C) 15 m/s (D) 20 m/s
|dy = vot + ½at2 → 45 m = 0 + ½(-10 m/s2)t2 ∴ t = 3 s |
|dx = vxt ∴ vx = dx/t = 30 m/3 s = 10 m/s |
13. A projectile is fired from the surface of the Earth with a speed of 200 m/s at an angle of 30° above the horizontal. If the ground is level, what is the maximum height reached by the projectile?
(A) 5 m (B) 10 m (C) 500 m (D) 1,000 m
|30o-60o-90o Δ (sides are 1-2-√3) ∴ vy = ½(200) = 100 m/s |
|v2 = vo2 + 2ad → (0)2 = (100 m/s)2 – 2(10 m/s2)d ∴ d = 500 m |
14. Vectors V1 and V2 have equal magnitudes and represent the velocities of an object at times t1 and t2, respectively.
[pic]
The direction of acceleration between time t1 and t2 is
(A) √ (B) © (C) ∇ (D) ∏
|in order for V1 to become V2, it needs to accelerate north and west ∴ ∏ |
Questions 15-16 A person on the west side of a river, which runs north to south, wishes to cross the 1-km wide river to a point directly across from his starting point. The river current is 3 km/hr and the boat's maximum speed is 5 km/hr.
15. What direction should the person direct the boat?
(A) due east (B) due north
(C) 37o north of east (D) 53o north of east
|5 km/hr, is the hypotenuse and 3 km/hr is the opposite side (south) ∴ |
|3-4-5 triangle with θ = 37o N of E. |
16. How long will it take to reach the dock?
(A) 1/5 hr (B) 1/4 hr (C) 1/2 hr (D) 1 hr
|the velocity east is 4 km/hr (3-4-5 triangle). Since the river is 1 km |
|wide, it will take ¼ hours |
17. A stationary observer on the ground sees a package falling with speed v1 at an angle to the vertical. At the same time, a pilot flying horizontally at constant speed relative to the ground sees the package fall vertically with speed v2. What is the speed of the pilot relative to the ground?
(A) v1 + v2 (B) v1 – v2
(C) v2 – v1 (D) (v12 – v22)½
|v2 is vertical, v is horizontal and v1 is hypotenuse |
|v12 = vpilot2 + v22 ∴ vpilot = (v12 – v22)½ |
18. How much time is a rock in the air if it is thrown horizontally off a building from a height h with a speed vo?
(A) (hvo)½ (B) h/vo (C) hvo/g (D) (2h/g)½
|dy = vot + ½at2 |
|h = 0 + ½gt2 ∴ t = (2gh)½ |
19. A spring-loaded dart gun generates velocity vo. What is the maximum height h reached by the dart if the gun is pointed at 30o above the horizontal?
(A) vo2/4g (B) vo/2g (C) vo2/8g (D) vo/g
|vy = vosin30 = vo/2 |
|vt2 = vo2 + 2ad |
|0 = (vo/2)2 + 2gh ∴ h = vo2/8g |
Practice Free Response
1. A physics student kicks a ball across the football field. It travels 50 m in 3.2 seconds.
a. What is the x-component of the initial velocity?
|vx = dx/t = 50 m/3.2 s = 16 m/s |
b. What is the y-component of the initial velocity?
|vt = vo + at → -vyo = vyo + at |
|∴ vyo = -½at = -½(-10 m/s2)(3.2 s) = 16 m/s |
c. What are the initial velocity's magnitude and direction?
|vo = (vxo2 + vyo2)½ = [(16 m/s)2 + (16 m/s)2]½ = 23 m/s |
|tanθ = vyo/vxo = (16 m/s)/(16 m/s) = 1 ∴ θ = 45o |
d. Graph the following.
dx vx ax
t t t
dy vy ay
t t t
2. A rescue plane, traveling horizontally at 70 m/s, drops supplies to mountain climbers on a rock ridge 235 m below.
[pic]
a. How much time is the package in the air?
|dy = vyot + ½at2 → -235 m = ½(-10 m/s2)t2 ∴ t = 6.9 s |
b. How far in advance, x, are the supplies dropped?
|x = vxt = (70 m/s)(6.9 s) = 483 m |
Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the climbers. What vertical velocity (up or down) is given to the supplies so that they reach the climbers?
[pic]
c. How much time is the package in the air?
|t = dx/vx = 425 m/70 m/s = 6.1 s |
d. What is the package's initial velocity?
|dy = vyot + ½at2 |
|-235 m = vyo(6.1 s) + ½(-10 m/s2)(6.1 s)2 ∴ vyo = -8.0 m/s |
3. Is it possible to hit a horizontal serve (without spin) that falls within the service box and clear the 0.90 m net?
[pic]
a. What is the minimum amount of time for the ball to be in the air?
|dy = vot + ½at2 → -2.50 m = ½(-10 m/s2)t2 ∴ t = 0.71 s |
b. What is the ball's maximum velocity?
|vx = 22.0 m/0.71 s = 31 m/s |
c. How long does it take for the ball to reach the net?
|t = dx/vx = 15.0 m/31 m/s = 0.48 s |
d. What is the ball's elevation when it reaches the net?
|dy = vot + ½at2 = ½(-10 m/s2)(0.48 s)2 = -1.2 m |
|∴ h = 2.50 m – 1.2 m = 1.3 m |
e. Does the ball clear the net?
|Yes, the net is 0.9 m, which is below the height of the ball. |
4. A boy aims a water-balloon slingshot at a second boy, who is hanging from a tree that is 10 m away. The balloon is launched with velocity, vo = 20 m/s at an angle θo = 15o at the same time the second boy drops from the tree.
[pic]
a. What are the horizontal and vertical components of the balloon's initial velocity?
|vxo = vocosθ = (20 m/s)cos(15) = 19 m/s |
|vyo = vosinθ = (20 m/s)sin(15) = 5 m/s |
b. How long does it take the balloon to reach the second boy?
|d = vt |
|10 m = (19 m/s)t ∴ t = 0.53 s |
c. How far did the second boy fall before he was hit by the balloon?
|d = vot + ½at2 |
|d = 0 + ½(-10 m/s2)(0.53 s)2 = -1.4 m |
d. What is the balloon's velocity when it strikes the second boy?
|vyt = vyo + at = 5 m/s + (-10 m/s2)(0.53 s) = 0.3 m/s |
|vt = (vxt2 + vyt2)½ = (192 + 0.32)½ = 19 m/s |
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