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Regents Physics – Midterm Review Free Response Name _________

1. The graph below shows the relationship between speed and elapsed time for a car moving in a straight line.

a. Determine the magnitude of the acceleration of the car. [1]

a = slope = 1.25 m/s/s

b. Calculate the total distance the car traveled during the time interval 4.0 seconds to 8.0 seconds. [Show all work, including the equation and substitution with units.] [2]

d = vit +1/2 at2 = (5 m/s)(4 s) + (1/2)(1.25 m/s/s)(4 s)2 = 30 m

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2. A 28-gram rubber stopper is attached to a string and whirled clockwise in a horizontal circle with a radius of 0.80 meter. The diagram in your answer booklet represents the motion of the rubber stopper. The stopper maintains a constant speed of 2.5 meters per second.

a. Calculate the magnitude of the centripetal acceleration of the stopper. [Show all work, including the equation and substitution with units.] [2]

ac = v2/r = (2.5 m/s)2 / (0.8 m) = 7.8 m/s2

b. On the diagram, draw an arrow showing the direction of the centripetal force acting on the stopper when it is at the position shown. [1]

3. A projectile is fired with an initial velocity of 120. meters per second at an angle, θ, above the horizontal. If the projectile’s initial horizontal speed is 55 meters per second. What is the measurement of angle θ?

θ = cos-1(55 m/s / 120 m/s) = 63o

4. Calculate the magnitude of the centripetal force acting on Earth as it orbits the Sun, assuming a circular orbit and an orbital speed of 3.00 × 104 meters per second. [Show all work, including the equation and substitution with units.] [2]

Fc = mv2/r = (5.98 kg x 1024)(3.00 x 104 m/s)2 / (1.5 x 1011 m) = 3.6 x 1022 N

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5. A student produced various elongations of a spring by applying a series of forces to the spring. The graph below represents the relationship between the applied force and the elongation of the spring.

a. Determine the spring constant of the spring. [1]

k = F / x = 20 N/m

b. Calculate the energy stored in the spring when the elongation is 0.30 meter. [Show all work, including the equation and substitution with units.] [2]

PEs = (1/2) kx2 = (1/2)(20 N/m)(0.3 m)2 = 0.9 J

6. A projectile is launched at an angle above the ground. The horizontal component of the projectile’s velocity, vx, is initially 40. meters per second. The vertical component of the projectile’s velocity, vy, is initially 30. meters per second. What are the components of the projectile’s velocity after 2.0 seconds of flight? [Neglect friction.][1]

vx = constant = 40 m/s

vy = viy + at = 30 m/s + (-9.81 m/s2) (2.0 s) = 10. m/s

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7. A girl rides her bicycle 1.40 kilometers west, 0.70 kilometer south, and 0.30 kilometer east in 12 minutes. The vector diagram in your answer booklet represents the girl’s first two displacements in sequence from point P. The scale used in the diagram is 1.0 centimeter 0.20 kilometer.

a. On the vector diagram, using a ruler and a protractor, construct the following vectors:

Starting at the arrowhead of the second displacement vector, draw a vector to represent the 0.30 kilometer east displacement. Label the vector with its magnitude. [1]

Draw the vector representing the resultant displacement of the girl for the entire bicycle trip and label the vector R. [1]

b. Calculate the girl’s average speed for the entire bicycle trip. [Show all work, including the equation and substitution with units.] [2]

v = d / t = (1.40 km + 0.70 km + 0.30 km) / 12 min = 0.2 km / min

c. Determine the magnitude of the girl’s resultant displacement for the entire bicycle trip, in kilometers. [1]

1.3 kmd. Determine the measure of the angle, in degrees, between the resultant

and the 1.40-kilometer displacement vector. [1]

27o

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8. A 30.4-newton force is used to slide a 40.0-newton crate a distance of 6.00 meters at constant speed along an incline to a vertical height of 3.00 meters.

a. Determine the total work done by the 30.4-newton force in sliding the crate along the incline. [1]W = Fdcosθ = (30.4 N)(6.00 m)cos0o = 182 J

b. Calculate the total increase in the gravitational potential energy of the crate after it has slid 6.00 meters along the incline. [Show all work, including the equation and substitution with units.] [2]

PE = mgh = (40 N)(3.0 m) = 120 J

c. State what happens to the kinetic energy of the crate as it slides along the incline. [1]

Kinetic energy is constant because the speed is constant

d. State what happens to the internal energy of the crate as it slides along the incline. [1]Internal energy of the crate is increasing as it slides along the

incline.

9. A skier starting from rest skis straight down a slope 50. meters long in 5.0 seconds. What is the magnitude of the acceleration of the skier? [Show all work, including the equation and substitution with units.] [2]

d = vit + (1/2)at2

50 m = (0 m/s)(5.0 s) + (1/2) a (5.0 s)2

a = 4 m/s2

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10. A machine launches a tennis ball at an angle of 45° with the horizontal, as shown. The ball has an initial vertical velocity of 9.0 meters per second and an initial horizontal velocity of 9.0 meters per second. The ball reaches its maximum height 0.92 second after its launch. [Neglect air resistance and assume the ball lands at the same height above the ground from which it was launched.]

a. What is the speed of the tennis ball as it leaves the launcher? [1]

vi=√vix2+v

iy2=√(9.0 m /s )2+(9 . 0m /s )2=13 m /s

b. What is the total horizontal distance traveled by the tennis ball during the entire time it is in the air? [Show all work, including the equation and substitution with units.] [2]

dx = vixt = (9.0 m/s) (0.92 s ∙ 2) = 17 m

11. A child kicks a ball with an initial velocity of 8.5 meters per second at an angle of 35º with the horizontal, as shown in the diagram. The ball has an initial vertical velocity of 4.9 meters per second and a total time of flight of 1.0 second. [Neglect air resistance.] What is the maximum height reached by the ball? [Show all work, including the equation and substitution with units.] [2]

v 2fy =v

iy2+2 ay d y

(0m /s )2=(4 . 9m /s )2+2 (−9 .81 m/s2)d y

d y=1.2m

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12. A soccer ball is kicked from point Pi at an angle above a horizontal field. The ball follows an ideal path before landing on the field at point Pf .

a. On the above diagram, draw an arrow to show the direction of the net force on the ball when it is at position X? [Neglect friction.] [1]

b. On the above diagram, draw an arrow to show the direction of acceleration on the ball when it is at position Y? [Neglect friction.] [1]

13. As shown in the diagram below, a student standing on the roof of a 50.0-meter-high building kicks a stone at a horizontal speed of 4.00 meters per second. How much time is required for the stone to reach the level ground below? [Neglect friction.][1]

d y=v iy t+12

a y t2

−50 m=(0 m /s )( t )+(12

)(−9. 81 m /s2) t 2

t=3 .2 s

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14. A machine fired several projectiles at the same angle, θ, above the horizontal. Each projectile was fired with a different initial velocity, vi. The graph below represents the relationship between the magnitude of the initial vertical velocity, viy, and the magnitude of the corresponding initial velocity, vi, of these projectiles. Determine the angle, θ, above the horizontal at which the projectiles were fired. [Show all work, including the equation and substitution with units.] [2]

viy=v i sinθ25 m /s=(40 m /s )sin θ

θ=sin−1 (25 m / s40 m /s

)=39o

15. Base your answer to the question on the graph below, which represents the relationship between velocity and time for a car moving along a straight line, and your knowledge of physics.

a. Determine the magnitude of the car’s acceleration during the first 6.0 seconds.[1]

a = slope = 2.5 m/s2

b. Determine the magnitude of the average velocity of the car from t = 6.0 seconds to t = 10. seconds.[1] 15m/s

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16. A truck, initially traveling at a speed of 22 meters per second, increases speed at a constant rate of 2.4 meters per second2 for 3.2 seconds. What is the total distance traveled by the truck during this 3.2-second time interval?[1]

d=v i t+12

at 2

d=(22 m /s )(3 .2 s )+12

(2 .4 m / s2 )(3 .2 s )2=83 m

17. What are the horizontal and vertical components of a 15-meters-per-second velocity at an angle of 60.° above the horizontal? [2]

vx=v cosθ=(15 m /s )cos60o=7 . 5m /svy=v sin θ=(15 m /s )sin 60o=13 m /s

18. A car traveling west in a straight line on a highway decreases its speed from 30.0 meters per second to 23.0 meters per second in 2.00 seconds. What is the car’s average acceleration during this time interval? [2]

a= Δvt

=23 . 0m /s−30 .0 m / s2. 00 s

=−3 .50 m /s2

19. If a 30-newton force is required to accelerate a 2-kilogram object at 10 meters per second2, over a level floor, what is the magnitude of the frictional force acting on the object? [1]

Fnet=Fapp−F fma=Fapp−F f

(2 kg )(10m/ s2 )=30 N−F f

F f=10 N

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20. A 10.-kilogram box, sliding to the right across a rough horizontal floor, accelerates at -2.0 meters per second2 due to the force of friction.

a. Calculate the magnitude of the net force acting on the box.[1]

Fnet=ma=(10 kg)(−2 . 0 m/ s2)Fnet=−20 N

b. Calculate the coefficient of kinetic friction between the box and the floor. [Show all work, including the equation and substitution with units.] [2]

F f=μFN

20 N=μ (10 kg)(9 .81 m /s2)μ=0 .2

21. A 1.2 × 103-kilogram automobile in motion strikes a 1.0 × 10-4-kilogram insect. As a result, the insect is accelerated at a rate of 1.0 × 102 m/s2. What is the magnitude of the force the insect exerts on the car? [Show all work, including the equation and substitution with units.] [2]Fnet=ma=(1.0⋅10−4 kg)(1 . 0⋅102 m /s2 )Fnet=0 . 01 N

22. A horizontal 20.-newton force is applied to a 5.0-kilogram box to push it across a rough, horizontal floor at a constant velocity of 3.0 meters per second to the right. Determine the magnitude of the force of friction acting on the box.[1] 20 N

23. A 4.0-kilogram object is accelerated at 3.0 meters per second2 north by an unbalanced force. What will be the acceleration on a 2.0 kg object if the same unbalanced force acting on it?[1] 6 m/s2

24. A 0.50-kilogram puck sliding on a horizontal shuffleboard court is slowed to rest by a frictional force of 1.2 newtons. What is the coefficient of kinetic friction between the puck and the surface of the shuffleboard court? [Show all work, including the equation and substitution with units.] [2]

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F f=μFN 1 .2 N=μ (0. 50 kg )(9 .81m / s2 ) μ=0 . 24

25. A student and the waxed skis he is wearing have a combined weight of 850 newtons. The skier travels down a snow-covered hill and then glides to the east across a snow-covered, horizontal surface. Calculate the magnitude of the force of friction acting on the skis as the skier glides across the snow-covered, horizontal surface. [Show all work, including the equation and substitution with units.] [2]F f=μFN F f=(0 .05 )(850 N )=42 .5 N

26. An ice skater applies a horizontal force to a 20.-kilogram block on frictionless, level ice, causing the block to accelerate uniformly at 1.4 meters per second2 to the right. After the skater stops pushing the block, it slides onto a region of ice that is covered with a thin layer of sand. The coefficient of kinetic friction between the block and the sand-covered ice is 0.28. Calculate the magnitude of the force of friction acting on the block as it slides over the sand-covered ice. [Show all work, including the equation and substitution with units.] [2]

F f=μFN F f=(0 .28 )(20 kg )(9 .81 m /s2 )=55 N

27. A 1.0 × 103 kilogram car travels at a constant speed of 20. meters per second around a horizontal circular track. The diameter of the track is 1.0 × 102 meters. What is the magnitude of the car’s centripetal acceleration? [Show all work, including the equation and substitution with units.] [2]ac=

v2

r=

(20 m /s )2

50 m=8m / s2

28. A 3.0-kilogram object is acted upon by an impulse having a magnitude of 15 newton•seconds. What is the magnitude of the object’s change in momentum due to this impulse?[1] ∆p = J = 15 N∙s

29. An 8.0-kilogram cart moving to the right at 4.0 meters per second about to make a head-on collision with a 4.0-kilogram cart moving to the left at 6.0 meters per second. After the collision, the 4.0-kilogram cart moves to the right at 3.0 meters per second. What is the velocity of the 8.0-kilogram cart after the collision? [Show all work, including the equation and substitution with units.] [2]m1 v1+m2v2=m1 v1 '+m2 v2 ' (8 . 0kg )(4 . 0m /s )+(4 . 0kg )(−6 .0m / s)=(8 . 0kg )v1 '+(4 . 0 kg )(3 .0m / s )v1 '=−0. 5m /s

30. A student on an amusement park ride moves in a circular path with a radius of 3.5 meters once every 8.9 seconds. What is the students average speed? [1]

v=dt=2 πr

T=

2π (3.5 m)8 .9 s

=2 .5m /s

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31. In an experiment, a 0.028-kilogram rubber stopper is attached to one end of a string. A student whirls the stopper overhead in a horizontal circle with a radius of 1.0 meter. The stopper completes 10 revolutions in 10 seconds.

a. Calculate the magnitude of the centripetal force on the whirling stopper. [Show all work, including the equation and substitution with units.] [2]

v=dt=2 πr

T=

2π (1m )1 s

=6 . 24 m / s Fc=mac=(0 .028 m/s )(6 . 24 m / s )2

1 m=1.1 N

b. Determine the speed of the whirling stopper.[1] 6.24m/s32. A 0.149-kilogram baseball, initially moving at 15 meters per second, is brought to rest in

0.040 second by a baseball glove on a catcher’s hand. What is the magnitude of the average force exerted on the ball by the glove?

F=ma=(0 .149 kg )(0 m /s−15m /s )

0 .040 s=−56 N

33. On a snow-covered road, a car with a mass of 1.1 × 103 kilograms collides head-on with a van having a mass of 2.5 × 103 kilograms traveling at 8.0 meters per second. As a result of the collision, the vehicles lock together and immediately come to rest. Calculate the speed of the car immediately before the collision. [Neglect friction.]

m1v1+m2 v2=m1v1 '+m2 v2 ' (1 .1⋅103kg )(v1)+(2 .5⋅103 kg )(8 .0 m /s )=0v1=18 m /s

34. A 1200-kilogram car moving at 12 meters per second collides with a 2300-kilogram car that is waiting at rest at a traffic light. After the collision, the cars lock together and slide. Eventually, the combined cars are brought to rest by a force of kinetic friction as the rubber tires slide across the dry, level, asphalt road surface. Calculate the speed of the locked-together cars immediately after the collision.

m1 v1+m2 v2=m1 v1 '+m2 v2 ' (1200 kg )(12 m/ s )+(2300 kg )(0 m /s )=(3500 kg )v 'v1=4 .1 m/s

35. Calculate the magnitude of the impulse applied to a 0.75-kilogram cart to change its velocity from 0.50 meter per second east to 2.00 meters per second east.

J= Δp=mΔv=(0 .75 kg )(2 .00 m / s−0. 50 m /s )=1 . 13 kg⋅m/ s

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36. When Earth and the Moon are separated by a distance of 3.84 × 108 meters, the magnitude of the gravitational force of attraction between them is 2.0 × 1020 newtons. What would be the magnitude of this gravitational force of attraction if Earth and the Moon were separated by a distance of 1.92 × 108 meters?

Since distance halved, Fg will quadruple. Fg = 8.0x1020 N

37. Io (pronounced “EYE oh”) is one of Jupiter’s moons discovered by Galileo. Io is slightly larger than Earth’s Moon. The mass of Io is 8.93 × 1022 kilograms and the mass of Jupiter is 1.90 × 1027 kilograms. The distance between the centers of Io and Jupiter is 4.22 × 108 meters.

a. Calculate the magnitude of the acceleration of Io due to the gravitational force exerted by Jupiter.[1]

g=Fg

m= 6 .35⋅1022 N

8. 93⋅1022kg=0 .711m / s2

b. Calculate the magnitude of the gravitational force of attraction that Jupiter exerts on Io.[1]

Fg=Gm1m2

r2 =(6 .67⋅10−11Nm2 /kg2 )(8 .93⋅1022 kg )(1.90⋅1027kg )

(4 . 22⋅108m )2 =6 . 35⋅1022 N

38. Two bowling balls, A and B, each having a mass of 7.00 kilograms, placed 2.00 meters apart. What is the magnitude of the gravitational force exerted by ball A on ball B?[1]

8.2x10-10 N

39. The weight of an object was determined at five different distances from the center of Earth. The results are shown in the table below. Position A represents results for the object at the surface of Earth. The mass of Earth is 5.98 x 1024 kg.

a. What is the mass of the object? [1] 100 kg

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b. At what distance from the center of Earth is the weight of the object approximately 28 newtons? [2]

Fg=Gm1 m2

r2

28 N=(6 .67⋅10−11 Nm2 /kg2)(5 .98⋅1024 kg )(100 kg )r 2

r=3 .77⋅107m

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