# Physics Practice Problems

The __________ of the particle at point A differ(s) as expressed in one coordinate system compared to the other, but the __________ from A to B is/are the same as expressed in both coordinate systems.

2. Vector

3. Vector

4. Scalar

6. Vector

6. Vector

7. Scalar

1. Position

2. Velocity

3. Displacement

4. Speed

5. Acceleration

6. Average velocity

7. Distance

a. Less than 1.0 s

b. 1.0 s

c. Greater than 1.0 s

d. Cannot be determined

a. Less than 1.0 s

b. 1.0 s

c. Greater than 1.0 s

d. Cannot be determined

a. 0 m/s

b. Between 0 m/s and 10 m/s

c. 10 m/s

d. Greater than 10 m/s

e. Cannot be determined

a. – x direction decreasing in speed

b. + x direction decreasing in speed

c. – x direction at a constant 20 m/s

d. – x direction increasing in speed

e. + x direction increasing in speed

a. The object will continue to move to the right, slowing down but never coming to a complete stop.

b. The object cannot have a negative acceleration and be moving to the right.

c. The object will slow down, eventually coming to a complete stop.

d. The object will slow down, momentarily stopping, then pick up speed moving to the left.

a. v = 0, a = 0

b. v = 9.8 m/s up, a = 0

c. v = 0, a = 9.8 m/s^2 up

d. v = 9.8 m/s down, a = 0

e. v = 0, a = 9.8 m/s^2 down

a. The velocity of the 1st rock increases faster than the velocity of the second.

b. The velocity of the second rock increases faster than the velocity of the first

c. Both velocities stay constant

d. Both increase at the same rate.

a. More than 40 km/hr

b. Less than 40 km/hr

c. Equal to 40 km/hr

d. Not enough info

a. The acceleration of the dropped ball is greater.

b. The acceleration changes during the motion, so you can’t predict the exact value when the 2 balls pass each other.

c. The accelerations are in opposite directions.

d. The acceleration of both balls is the same.

e. The acceleration of the ball thrown upward is greater

a. 20 m/s = (v+v0)/2

b. v = (20 m/s) – gt

c. v^2 = (-20 m/s)^2 – 2g(y-y0)

d. y = y0 + (-20 m/s)t – (1/2)gt^2

e. All of the above

Type: 3.1, 3.2

a. Align the adjacent side of a right triangle with the vector and the hypotenuse along a coordinate direction with ? as the included angle.

b. Align the hypotenuse of a right triangle with the vector and an adjacent side along a coordinate direction with ? as the included angle.

c. Align the opposite side of a right triangle with the vector and the hypotenuse along a coordinate direction with ? as the included angle.

d. Align the hypotenuse of a right triangle with the vector and the opposite side along a coordinate direction with ? as the included angle

a. 0

b. 18

c. 37

d. 64

e. 100

a. less than or equal to the magnitude of the vector

b. equal to the magnitude of the vector

c. greater than or equal to the magnitude of the vector

d. less than, equal to, or greater than the magnitude of the vector

a. as soon as it leaves the barrel

b. after air friction reduces its speed

c. not at all if air resistance is ignored

a. angled upward in the direction of motion

b. directly down

c. angled down in the opposite direction of motion

a. The dropped ball

b. The thrown ball

c. Neither – they both have the same speed on impact

d. It depends on how hard the ball was thrown.

a. the magnitude of the acceleration is 0

b. the magnitude of the velocity is 0

c. the magnitude of the velocity is slowest

d. more than 1 of the above is true

e. none of the above

a. The bullet will hit the monkey because both the monkey and the bullet are falling downward at the same rate due to gravity.

b. The bullet will miss the monkey because the monkey falls down while the bullet speeds straightforward.

c. It depends on how far the hunter is from the monkey.

d. The bullet will miss the monkey because although both the monkey and the bullet are falling downward due to gravity, the monkey is falling faster.

a. The projectile has the same x velocity at any point on its path.

b. The acceleration of the projectile is positive and decreasing when the projectile is moving upwards, zero at the top, and increasingly negative as the projectile descends.

c. The acceleration of the projectile is a constant negative value.

d. The y component of the velocity of the projectile is zero at the highest point of the projectile’s path.

e. The velocity at the highest point is zero.

a. there must be no forces applied to the car

b. the net force applied to the car is directed to the left

c. the net force applied to the car is 0

d. there is exactly one force applied to the car

a. there are no forces at all acting on it

b. the net force acting on it is 0

c. the net force acting on it is constant

d. there is only 1 force acting on it

(Select the most general response)

a. there are no forces at all acting on it

b. the net force acting on it is 0

c. the net force acting on it is constant in magnitude and direction

d. there is only 1 force acting on it

a. The net force acting on it is zero.

b. The net force acting on it is constant and nonzero.

c. There are no forces at all acting on it.

d. There is only one force acting on it.

a. There is exactly one force applied to the block.

b. The net force applied to the block is directed to the left.

c. The net force applied to the block is zero.

d. There must be no forces at all applied to the block.

a. It must be moving to the left.

b. It must be moving to the right

c. It must be at rest

d. It could be moving to the left, moving to the right, or be instantaneously at rest.

a. continuously changing direction

b. moving at constant velocity

c. moving with a constant nonzero acceleration

d. moving with continuously increasing acceleration

a. cannot have a magnitude equal to 5N

b. cannot have a magnitude equal to 10 N

c. cannot have the same direction as the force w/ magnitude 10N

d. must have a magnitude equal to 10N

a. the table

b. gravity

c. inertia

earth

book

upward

table

book

downward

(To start moving forward, sprinters push backward on the starting blocks with their feet. As a reaction, the blocks push forward on their feet with a force of the same magnitude. This external force accelerates the sprinter forward.)

a. the blocks

b. the sprinter

a. greater in magnitude than, and in the opposite direction from, the force on the earth due to the moon.

b. greater in magnitude than, and in the same direction as, the force on the earth due to the moon.

c. equal in magnitude to, and in the opposite direction from, the force on the earth due to the moon.

d. equal in magnitude to, and in the same direction as, the force on the earth due to the moon.

e. smaller in magnitude than, and in the opposite direction from, the force on the earth due to the moon.

f. smaller in magnitude than, and in the same direction as, the force on the earth due to the moon.

a. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase.

b. The magnitude of the normal force is equal to the magnitude of the weight of the suitcase minus the magnitude of the force of the pull.

c. The magnitude of the normal force is equal to the sum of the magnitude of the force of the pull and the magnitude of the suitcase’s weight.

d. The magnitude of the normal force is greater than the magnitude of the weight of the suitcase.

a. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight.

b. The magnitude of the normal force is equal to the magnitude of the suitcase’s weight minus the magnitude of the man’s weight.

c. The magnitude of the normal force is equal to the sum of the magnitude of the man’s weight and the magnitude of the suitcase’s weight.

d. The magnitude of the normal force is less than the magnitude of the suitcase’s weight.

a. The frictional force points up the incline

b. The frictional force points down the incline

c. The frictional force is zero.

a. The frictional force points up the incline

b. The frictional force points down the incline

c. The frictional force is zero.

F of 2 physicists (is less than/equals/is greater than) F of friction.

a. No direction.. The net force is 0

b. Straight down because of gravity

c. Horizontal and to the left

d. Straight up (the normal force)

e. Horizontal and to the right

a. The friction force exerted by the road on you

b. The force exerted by the car on you

c. The friction force exerted by the car on the road

d. The normal force exerted by the road on you.

a. the truck offers no resistance because its brakes are off.

b. he is heavier in some respects than the truck.

c. he exerts a greater force on the truck than the truck exerts back on him.

d. the ground exerts a greater friction force on Matt than it does on the truck.

e. he is stronger than the truck.

a. doesn’t change

b. increases until the rope is straight

c. decreases as the pack rises until the rope is straight across

d. increases but the rope always sags where the pack hangs

a. The force the paddle exerts on the water.

b. The force the man exerts on the paddle.

c. The force the water exerts on the paddle.

d. The motion of the water itself.

a. is moving downward at constant velocity

b. is accelerating upward

c. is moving upward at constant velocity

d. is accelerating downward

e. is standing still

a. equal to the weight of the skier

b. less than the weight of the skier

c. greater than the weight of the skier

a. The force of the golf club acting on the ball.

b. The force of gravity acting on the ball.

c. The force of the ball moving forward through the air.

d. All of the above

e. Both A and C

a. is brought to rest and then accelerates in the direction of the second force.

b. decelerates gradually to rest.

c. is brought to rest rapidly.

d. continues at the velocity it had before the second force was applied.

a. you exert a force on the box, but the box does not exert a force on you.

b. the force that the box exerts on you is greater than the force you exert on the box.

c. the force you exert on the box is equal to the force of the box pushing back on you.

d. the box is so heavy it exerts a force on you, but you do not exert a force on the box.

e. the force you exert on the box is greater than the force of the box pushing back on you

a. 20N to the left

b. 25N to the left

c. 25N to the right

d. 20N to the right

e. None of the above – the crate starts to move

a. exactly 950 N

b. less than 950 N

c. more than 950 N

Since the elevator is not accelerating, the reading on the scale is the same as in the video.

a. less than 18 N but greater than 0 N

b. 18 N

c. More than 18 N

d. 0 N

a. The speed of the cylinder has decreased

b. The speed of the cylinder has increased

c. The magnitude of the acceleration of the cylinder has decreased

d. The magnitude of the acceleration of the cylinder has increased

e. The speed and the acceleration of the cylinder have not changed.

a. V

b. V/2

c. 2V

d. sqrt(2) * V

e. V / sqrt(2)

a. V

b. V/6

c. 6V

d. sqrt(6) * V

e. V / sqrt(6)

a. v(s) = 1/4 * v(L)

b. v(s) = 1/2 * v(L)

c. v(s) = v(L)

d. v(s) = 2*v(L)

e. v(s) = 4*v(L)

a. F1 = 1/3 * F2

b. F1 = 3/4 * F2

c. F1 = F2

d. F1 = 3*F2

e. F1 = 27F2

The closer together two masses are, the stronger is the gravitational attraction between them. Thus, the mass at the origin is more strongly attracted to the mass at x1 = -110 cm than it is to the mass at x2 = 450 cm . Thus, the net force on the mass at the origin is in the -x direction.

a. +x direction

b. -x direction

Newton’s law of universal gravitation gives the magnitude of the gravitational force but does not distinguish between which object exerts the force and which object experiences the force.

a. The earth does not exert any gravitational force on the sun.

b. The earth exerts some force on the sun, but less than 3.53×1022N because the earth, which is exerting the force, is so much less massive than the sun.

c. The earth exerts 3.53×1022N of force on the sun, exactly the same amount of force the sun exerts on the earth found in Part A.

d. The earth exerts more than 3.53×1022N of force on the sun because the sun, which is experiencing the force, is so much more massive than the earth.

a. Reduce the mass of the earth to 1/4 its normal value.

b. Reduce the mass of the sun to 1/4 its normal value

c. Reduce the mass of the earth to 1/2 its normal value and the sun to one half its normal value.

d. Increase the separation between the earth and the sun to 4x its normal value.

a. The force is toward the sun

b. The force is toward the earth.

c. There is no net force because neither the sun nor the earth attracts the probe gravitationally at the midpoint.

d. There is no net force because the gravitational attractions on the probe due to the sun and the earth are equal in size but point in opposite directions, so they cancel each other out.

a. 12 lb

b. 0.34 g

c. 120 kg

d. 1600 kN

e. 0.34 m

f. 411 m

a. 12 lb

b. 0.34 g

c. 120 kg

d. 1600 kN

e. 0.34 m

f. 411 m

a. mass only

b. weight only

c. both mass and weight

d. neither mass nor weight

a. mass increases, weight decreases

b. mass decreases; weight increases

c. mass increases; weight increases

d. mass increases; weight remains the same

e. mass remains the same; weight decreases

f. mass remains the same; weight increases

g. mass remains the same; weight remains the same

The speed v of a satellite of mass m orbiting a distance r from the center of a larger object of mass M is given by the relationship

v=?(GM/r)

a. At a distance less than r

b. At a distance equal to r

c. At a distance greater than r

(Kepler’s)

a. Tm is less than Te

b. Tm = Te

c. Tm is greater than Te

a. The door is exerting a rightward force on you

b. Centrifugal force is pushing you into the door

c. Both of the above

d. None of the above

a. Velocity, acceleration, net force

b. velocity, acceleration

c. velocity, net force

d. acceleration

e. acceleration, net force

a. up at larger angle

b. up at smaller angle

c. straight from where the tube ends

d. down at small angle

e. down at larger angle

a. The net force on the car is zero.

b. Both the acceleration and net force on the car point outward.

c. The car’s acceleration is zero.

d. Both the acceleration and net force on the car point inward.

e. If there is no friction, the acceleration is outward.

a. A little after the bottom of the circle when the ball is climbing.

b. At the bottom of the circle.

c. Nowhere; the cord is stretched the same amount at all points.

d. A little before the bottom of the circle when the ball is descending quickly.

e. At the top of the circle.

a. Both the same; the Earth.

b. The Moon on the Earth; the Moon.

c. The Earth on the Moon; the Moon.

d. The Earth on the Moon; the Earth.

e. The Moon on the Earth; the Earth.

f. Both the same; the Moon.

a. The lighter satellite moves twice as fast as the heavier one.

b. The heavier satellite moves twice as fast as the lighter one.

c. The two satellites have the same speed.

d. The ratio of their speeds depends on the orbital radius.