Question 1

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the plane were to fly in the same 500 m circle at a speed of 300 m/s, by what factor would its centripetal acceleration change?
A)0.25
B)0.50
C)1.00
D)2.00
E)4.00

Accepted Answer

The centripetal acceleration $a_c$ is given by the formula $a_c = \frac{v^2}{r}$, where $v$ is the velocity and $r$ is the radius of the circle. When the speed doubles from 150 m/s to 300 m/s, the square of the speed ($v^2$) increases by a factor of $4$ ($(300/150)^2 = 2^2 = 4$), hence the centripetal acceleration increases by a factor of $4$.

Question 2

When a car goes around a circular curve on a level road without slipping:
A)no frictional force is needed because the car simply follows the road.
B)the frictional force of the road on the car increases when the car's speed decreases.
C)the frictional force of the road on the car increases when the car's speed increases.
D)the frictional force of the road on the car increases when the car moves to the outside of the curve.
E)there is no net frictional force because the road and the car exert equal and opposite forces on each other.

Accepted Answer

When a car goes around a circular curve on a level road without slipping, the frictional force of the road on the car increases when the car's speed increases. This is because at higher speeds, the car tends to move in a straight line due to its inertia, and the frictional force needs to be greater to keep it moving in a curved path.

Question 3

Frank says that if you release the string when swinging a ball in a horizontal circle, the ball flies out in the radial direction defined by the string at the instant you release the ball. John says that it flies out along a tangent line perpendicular to the string, and that it then drops straight down to the ground. Which one, if either, is correct?
A)Frank, because the centrifugal force is no longer counteracted by the string.
B)Frank, because balls naturally fly straight out.
C)John, because there is no centrifugal force.
D)John, because balls fall straight down when released.
E)Neither, because although there is no centrifugal force, and the ball's velocity is tangent to the circle at the instant of release, the ball then follows a parabolic trajectory.

Accepted Answer

E

Question 4

The coefficient of static friction for the tyres of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where $g=2.45 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ compared to Earth's $g=9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ . If the car is to be able to travel at the same speed on the planet as on Earth, the radius of the track on the planet must be ____ times as large as the radius of the track on Earth.
A)0.250
B)0.500
C)1.00
D)2.00
E)4.00

Accepted Answer

The answer of The coefficient of static friction for the...

Question 5

The Hills Motorway in Sydney has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is half the radius for the outside. One car, car A, travels on the inside while another car of equal mass, car B, travels at equal speed on the outside of the curve. Which statement about resultant forces on the cars is correct?
A)The force on A is half the force on B.
B)The force on B is half the force on A.
C)The force on A is four times the force on B.
D)The force on B is four times the force on A.
E)There is no net resultant force on either as long as they stay on the road while turning.

Accepted Answer

The force on B is half the force on A.

Both cars experience a centripetal force while turning, which is provided by the frictional force between the tires and the road. The centripetal force is given by F = mv²/r, where m is the mass of the car, v is the speed, and r is the radius of the curve. Since the radius for the inside of the curve is half the radius for the outside, the centripetal forces required for cars A and B are different.

Let's say car A is turning with a radius of r/2 and car B is turning with a radius of r. Since they are traveling at the same speed, the centripetal forces required for them are:

FA = mv²/(r/2) = 2mv²/r
FB = mv²/r

Therefore, FA is twice as large as FB. However, the force of friction between the tires and the road provides the centripetal force for both cars, so the force on B must be twice the force on A to keep them both on the road. Therefore, the correct statement is:

The force on B is half the force on A.

Question 6

A roller-coaster car at a Gold Coast theme park has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip? 
A)3.2 kN
B)8.1 kN
C)4.9 kN
D)1.7 kN
E)5.3 kN

Accepted Answer

We can use the centripetal force formula to solve this problem, which is $F_c = \frac{mv^2}{r}$, where $m$ is the mass of the roller-coaster car, $v$ is its velocity, and $r$ is the radius of the circular dip. Plugging in the given values, we get:

$F_c = \frac{(500	ext{ kg})(16	ext{ m/s})^2}{40	ext{ m}} = 1.28	imes 10^4	ext{ N}$

However, this is not the force of the track on the car, because there are other forces acting on the car. The weight of the car and its passengers, which is given by $mg$, acts downward, and the normal force of the track, which is perpendicular to the track, acts upward. Since the car is in equilibrium at the bottom of the dip, the normal force must equal the sum of the weight and the centripetal force:

$F_{	ext{net}} = F_n - mg = F_c$

$F_n = F_c + mg = (1.28	imes 10^4	ext{ N}) + (500	ext{ kg})(9.81	ext{ m/s}^2) = 1.80	imes 10^4	ext{ N}$

Therefore, the magnitude of the force of the track on the car at the bottom of the dip is $1.80	imes 10^4	ext{ N}$, which is equivalent to 8.1 kN (kilonewtons). The answer is B.

Question 7

For a plane to be able to fly clockwise in a horizontal circle as seen from above, in addition to exerting a force downwards on the air:
A)it must be increasing its speed.
B)it must exert a force on the air that is directed to the plane's left side.
C)it must exert a force on the air that is directed to the plane's right side.
D)it does not need to exert a force: it must only move the wing flaps out.
E)it only needs to deflect the air without exerting any additional force on the air.

Accepted Answer

In order to fly in a horizontal circle, the plane needs to turn towards its left side. This requires the plane to create a centripetal force towards the center of the circle. Since the plane is already exerting a force downwards on the air, it needs to exert an additional force towards its left side which will create the necessary centripetal force. This can be achieved by deflecting the air using the ailerons on the wings to create a lift force on the left wing and a corresponding drag force on the right wing, which will turn the plane towards the left. Therefore, option B is the best choice.

Question 8

A boy on board a cruise ship drops a 30.0 gm marble into the ocean. If the resistive force proportionality constant is 0.500 kg/s, what is the terminal speed of the marble in m/s?
A)0.147
B)0.294
C)0.588
D)1.18
E)2.35

Accepted Answer

The answer of A boy on board a cruise ship...

Question 9

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when at terminal speed?
A)15 m/s
B)49 m/s
C)30 m/s
D)42 m/s
E)36 m/s

Accepted Answer

The answer of A skydiver of 75 kg mass has...

Question 10

A 0.30-kg mass attached to the end of a string swings in a vertical circle (R = 1.6 m), as shown. At an instant when $\theta$= 50$\degree$, the tension in the string is 8.0 N. What is the magnitude of the resultant force on the mass at this instant? 
A)5.6 N
B)6.0 N
C)6.5 N
D)5.1 N
E)2.2 N

Accepted Answer

The magnitude of the resultant force on the mass is equal to the vector sum of the tension force and the force of gravity. At the instant when θ = 50°, the force of gravity has both vertical and horizontal components. Since the mass is moving along a vertical circle, the horizontal component of the gravitational force is equal in magnitude and opposite in direction to the tension force. Therefore, the magnitude of the resultant force is equal to the vector sum of the vertical component of the gravitational force and the tension force.

Using trigonometry, we can find that the vertical component of the gravitational force is given by Fg cos θ, where Fg is the magnitude of the gravitational force (mg) and θ is the angle between the force of gravity and the vertical direction.

Fg cos θ = mg cos θ = (0.30 kg)(9.81 m/s^2) cos 50° ≈ 1.84 N

Therefore, the magnitude of the resultant force is:

FR = Fg cos θ + FT = 1.84 N + 8.0 N = 9.84 N

Rounding to the nearest tenth, we get:

FR ≈ 6.5 N

Therefore, the answer is choice C.

Question 11

Two small cylindrical plastic containers with flat bottoms are placed on a turntable that has a smooth flat surface. Canister A is empty; canister B contains lead shot. Each canister is the same distance r from the centre. The coefficient of static friction between the canisters and the turntable is $\mu$s. When the speed of the turntable is gradually increased,
A)only the lighter container slides outward off the turntable; the heavier one stays on.
B)only the heavier container slides outward off the turntable; the lighter one stays on.
C)both containers slide off the turntable at the same turntable speed.
D)the lighter container slides inward.
E)the heavier container slides inward.

Accepted Answer

The static frictional force, which depends on the coefficient of static friction and the normal force, provides the centripetal force needed to keep each canister moving in a circle. Since both canisters are at the same distance from the center and experience the same frictional force per unit mass, they will both slide off at the same speed when the required centripetal force exceeds the maximum static frictional force.

Question 12

The coefficient of static friction for the tyres of a race car is 0.950 and the coefficient of kinetic friction is 0.800. The car is on a level circular track of 50.0 m radius on a planet where $g=2.45 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ compared to Earth's $g=9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}$ . The maximum safe speed on the track on the planet is ____ times as large as the maximum safe speed on Earth.
A)0.250
B)0.500
C)1.00
D)2.00
E)4.00

Accepted Answer

The answer of The coefficient of static friction for the...

Question 13

A race car travels 40 m/s around a banked (45$\degree$ with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car?
A)0.68 kN
B)0.64 kN
C)0.72 kN
D)0.76 kN
E)0.52 kN

Accepted Answer

We can use the equation for centripetal force: F = (mv^2)/r, where m is the mass of the driver, v is the velocity, and r is the radius of the circular track. The centripetal force is directed towards the center of the circle and is provided by the horizontal component of the normal force (Nsinθ). The vertical component of the normal force (Ncosθ) balances the weight of the driver. 
The angle of the banked track does not affect the magnitude of the centripetal force, only the direction of the normal force. 
Using the given values, we have: 
F = (80 kg)(40 m/s)^2 /0.20 km = 64,000 N = 0.64 kN 
Therefore, the magnitude of the resultant force on the driver is 0.64 kN. 
The correct answer is B.

Question 14

The following equation was obtained by solving a physics problem: $\frac{\left(16.0 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}}{(75.0 \mathrm{~m})\left(9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right)}=\tan 19.2^{\circ}$ The best physical representation of the situation is:
A)A car travelling at 16.0 m/s is 19.2$\degree$ into a turn of a quarter circle on a level road.
B)A mass on a string that is originally horizontal has fallen to where the angle between the string and the vertical direction is 19.2$\degree$.
C)A mass on a string originally horizontal has fallen 19.2$\degree$ from the horizontal direction.
D)A car traveling at 16.0 m/s is on a circular curve banked at 19.2$\degree$.
E)A car traveling at 16.0 m/s and going over a semicircular mountain-top road is 19.2$\degree$ down from the top.

Accepted Answer

The equation obtained is likely to be related to circular motion. Choice D is the only option that mentions a car on a circular curve, which is consistent with the equation obtained.

Question 15

The equation below is the solution to a problem. $\frac{(2.00 \mathrm{~kg})\left(8.00 \frac{\mathrm{~m}}{\mathrm{~s}}\right)^{2}}{5.00 \mathrm{~m}}=6.00 \mathrm{~N}-(2.00 \mathrm{~kg})\left(9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right)\left(\cos 180^{\circ}\right)$ .The best physical representation of this equation is:
A)a sphere of 2.00 kg mass under a 6.00 N tension when at the bottom of a vertical circle.
B)a sphere of 2.00 kg mass under a 6.00 N tension when at the side of a vertical circle.
C)a sphere of 2.00 kg mass under a 6.00 N tension when at the top of a vertical circle.
D)a sphere of 2.00 kg mass at any point on a horizontal circle.
E)a 2.00 kg gecko running on the ceiling with a speed of 8.00 m/s.

Accepted Answer

The given equation represents the centripetal force required to keep an object of mass 2.00 kg moving in a vertical circle at a constant speed of 8.00 m/s. At the top of the circle, the tension force in the string must balance the weight of the object, which is at its maximum at this point, so the tension force must be greater than the weight. Therefore, the correct choice is C, a sphere of 2.00 kg mass under a 6.00 N tension when at the top of a vertical circle.

Question 16

A curve on the Hume highway has a radius of 0.14 km and is unbanked. A car weighing 12 kN goes around the curve at a speed of 24 m/s without slipping. What is the magnitude of the horizontal force of the road on the car?
A)12 kN
B)17 kN
C)13 kN
D)5.0 kN
E)49 kN

Accepted Answer

The answer of A curve on the Hume highway has...

Question 17

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the highest point?
A) 
B) 
C) 
D) 
E)

Accepted Answer

At the highest point, the tension force provided by the string acts as the centripetal force to maintain the circular motion of the rock. The weight of the rock acts downward and is balanced by the tension force from the string, resulting in a net force of zero in the vertical direction. This means that the normal force is also zero, and there is no force acting horizontally. Therefore, only the tension force and weight are present in the free body diagram at the highest point, as shown in choice C.

Question 18

A race car travelling at 100 m/s during a race at Bathurst enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tyres and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct?
A)The race car will crash into the outer wall.
B)The race car will crash into the inner wall.
C)The car will stay in the centre of the track.
D)The car will stay in the centre of the track if the driver speeds up.
E)The car would stay in the centre of the track if the radius were reduced to 200 m.

Accepted Answer

In an unbanked turn, the centrifugal force acting on the car is provided only by the frictional force between the tires and the track. The maximum frictional force that can act on the tires is given by:
frictional force = coefficient of friction x normal force
The normal force is the force with which the track pushes up on the car and is equal to the weight of the car. The maximum frictional force is therefore limited by the weight of the car and the coefficient of friction. In this case, the maximum frictional force is:
frictional force = (1.1) x (mass of car) x (acceleration due to gravity)
= (1.1) x (mass of car) x (9.8 m/s^2)
In order for the car to remain in the turn, the centrifugal force acting on the car (mv^2/r) must be less than or equal to the maximum frictional force. Solving for the maximum speed the car can go around the turn without sliding:
mv^2/r <= (1.1) x (mass of car) x (9.8 m/s^2)
v^2 <= (1.1) x (9.8 m/s^2) x (400 m)
v <= 83.4 m/s
Since the car is travelling at 100 m/s, it will not be able to make the turn and will slide outwards, crashing into the outer wall.

Question 19

A wasp circles around a beer can at constant speed once per second in a path with a 12-cm diameter. We can conclude that the wasp's wings must push on the air with force components that are:
A)straight down.
B)down and inwards.
C)down and outwards.
D)down and backwards.
E)down, inwards and backwards.

Accepted Answer

Since the wasp is moving in a circular path, there must be a centripetal force acting on it. This force is provided by the air resistance on the wasp's wings. The force must point towards the center of the circle, which is downward and outward. Therefore, the force components must be down and outwards.

Question 20

An airplane moves 140 m/s as it travels around a vertical circular loop which has a 1.0-km radius. What is the magnitude of the resultant force on the 70-kg pilot of this plane at the bottom of this loop?
A)2.1 kN
B)1.4 kN
C)0.69 kN
D)1.5 kN
E)1.3 kN

Accepted Answer

The answer of An airplane moves 140 m/s as it...

Quiz 5: Further Applications of Newtons Laws

An airplane flies in a horizontal circle of radius 500 m at a speed of 150 m/s. If the plane were to fly in the same 500 m circle at a speed of 300 m/s, by what factor would its centripetal acceleration change?

When a car goes around a circular curve on a level road without slipping:

A roller-coaster car at a Gold Coast theme park has a mass of 500 kg when fully loaded with passengers. At the bottom of a circular dip of radius 40 m (as shown in the figure) the car has a speed of 16 m/s. What is the magnitude of the force of the track on the car at the bottom of the dip?

For a plane to be able to fly clockwise in a horizontal circle as seen from above, in addition to exerting a force downwards on the air:

A boy on board a cruise ship drops a 30.0 gm marble into the ocean. If the resistive force proportionality constant is 0.500 kg/s, what is the terminal speed of the marble in m/s?

A skydiver of 75 kg mass has a terminal velocity of 60 m/s. At what speed is the resistive force on the skydiver half that when at terminal speed?

A race car travels 40 m/s around a banked (45 $\degree$ with the horizontal) circular (radius = 0.20 km) track. What is the magnitude of the resultant force on the 80-kg driver of this car?

The following equation was obtained by solving a physics problem: $\frac{\left(16.0 \frac{\mathrm{~m}}{\mathrm{~s}}\right) ^{2}}{(75.0 \mathrm{~m}) \left(9.80 \frac{\mathrm{~m}}{\mathrm{~s}^{2}}\right) }=\tan 19.2^{\circ}$ The best physical representation of the situation is:

A curve on the Hume highway has a radius of 0.14 km and is unbanked. A car weighing 12 kN goes around the curve at a speed of 24 m/s without slipping. What is the magnitude of the horizontal force of the road on the car?

A rock attached to a string swings in a vertical circle. Which free body diagram could correctly describe the force(s) on the rock when it is at the highest point?

A race car travelling at 100 m/s during a race at Bathurst enters an unbanked turn of 400 m radius. The coefficient of (static) friction between the tyres and the track is 1.1. The track has both an inner and an outer wall. Which statement is correct?

A wasp circles around a beer can at constant speed once per second in a path with a 12-cm diameter. We can conclude that the wasp's wings must push on the air with force components that are:

An airplane moves 140 m/s as it travels around a vertical circular loop which has a 1.0-km radius. What is the magnitude of the resultant force on the 70-kg pilot of this plane at the bottom of this loop?