2 Kinematics

Question 1

Define distance. Is it a scalar or a vector quantity?

Answer 1

Distance represents the total length of the path an object travels. It is a scalar quantity.

Question 2

Define displacement. Is it a scalar or a vector quantity?

Answer 2

Displacement is the distance moved in a specified direction from a reference point. It is a vector quantity.

Question 3

Define speed. Is it a scalar or a vector quantity? How can average speed be found?

Answer 3

The instantaneous speed of an object is the rate of change of distance travelled with respect to time.
(v = dx/dt in ms^-1) It is a scalar quantity. Average speed equals the total distance travelled by an object/total time taken.

Question 4

Define velocity. Is it a scalar or a vector quantity? How can average velocity be found?

Answer 4

The instantaneous velocity of an object is the rate of change of displacement with respect to time.
(v = ds/dt in ms^-1) It is a vector quantity. Average velocity equals the change in displacement/total time taken.

Question 5

Define acceleration. Is it a scalar or a vector quantity? How can average acceleration be found?

Answer 5

The instantaneous acceleration of an object is the rate of change of velocity with respect to time.
(a = dv/dt in ms^-2) It is a vector quantity. Average acceleration equals the change in velocity/total time taken.

Question 6

Which two factors affect the sign of acceleration? When is acceleration positive? When is acceleration negative?

Answer 6

The sign of acceleration depends on whether the object is:
1) speeding up or slowing down? (accelerating or decelerating)
2) moving in the positive or negative direction? (displacement)

Acceleration is positive when the object is accelerating in the direction of positive displacement or decelerating in the direction of negative displacement. Acceleration is negative when the object is accelerating in the direction of negative displacement or decelerating in the direction of positive displacement.

Question 7

Using velocity and acceleration, how do we decide if an object is slowing up or speeding up?

Answer 7

When velocity and acceleration are in the same direction, the object is speeding up. When velocity and acceleration are in opposite directions, the object is slowing down.

Question 8

Sketch the (i) displacement-time, (ii) velocity-time and (iii) acceleration time graphs of an object travelling at increasing speed and constant acceleration. Give an example of when this occurs.

Answer 8

Releasing a ball at rest at a height
Graph on pg 6 of notes

Question 9

Sketch the (i) displacement-time, (ii) velocity-time and (iii) acceleration time graphs of an object travelling at decreasing speed and constant acceleration. Give an example of when this occurs.

Answer 9

Throwing a ball upwards

Question 10

What is the significance of the gradients of the (i) displacement-time (s-t), (ii) velocity-time (v-t) and (iii) acceleration-time (a-t) graphs of an object?

Answer 10

(i) Instantaneous velocity
(ii) Instantaneous acceleration
(iii) No significance

Question 11

What is the significance of the area under the graph of the (i) displacement-time (s-t), (ii) velocity-time (v-t) and (iii) acceleration-time (a-t) graphs of an object?

Answer 11

(i) No significance
(ii) Change in displacement
(iii) Change in velocity

Question 12

What criteria must be fulfilled before the equations of motion can be used?

Answer 12

One dimensional motion
Constant acceleration (same magnitude and direction)

Question 13

State the four equations of motions

Answer 13

v = u+at

s = 1/2(u+v)t

s = ut+1/2at^2

v^2 = u^2+2as

Note: When (iv) is multiplied by m/2, it gives the equation that the change in K.E numerically equals to the work done by a constant F.

Question 14

What is free fall?

Answer 14

When an object is falling in the absence of air resistance, it is undergoing free fall.

Question 15

What does an object experience when it is being thrown upwards or downwards or released from rest in the absence of air resistance?

Answer 15

The object experience a constant acceleration directed downwards (towards the centre of the earth) with a magnitude of g = 9.81ms^-1 when it is being thrown upwards or downwards or released from rest. This is known as the acceleration due to free fall and it is assumed to be a constant value near the Earth’s surface.

Question 16

What happens to an object when it moves through a fluid (presence of air resistance)?

Answer 16

When an object moves through a fluid (liquid/gas), it experiences a drag force/air resistance which acts opposite in direction to velocity.

Question 17

Under non-turbulent conditions, how is the drag force related to velocity?

Answer 17

The drag force is proportional to the instantaneous velocity. (FD = kv, where k is a constant)

Question 18

At higher velocities, how is the drag force related to velocity?

Answer 18

When turbulence sets in, the drag force is proportional to the square of the velocity. (Fd = k’v^2, where k’ is a constant)

Question 19

Describe the motion of an object released from rest in a uniform gravitational field where there is air resistance.

Answer 19

At the point of release, velocity is 0 and the only force acting on the object at this instant is the gravitational force. The object falls with an acceleration of g. As the body gains velocity, air resistance which is dependent on velocity starts to increase. The resultant force on the body = mg-Fd = ma. The body is still accelerating but acceleration is now less than g (a = g-Fd/m). Eventually, drag force increases to a point where is it equal in magnitude to the weight of the body and resultant force equals 0, acceleration of the body equals 0 and the body attains a constant velocity which is called the terminal velocity.

Question 20

Why is air resistance negligible for heavier objects?

Answer 20

a = g-Fd/m where g almost equals a.
amount of air resistance an object experience depends on its speed, its cross-sectional area, its shape and the density of the air.

Question 21

Sketch the graphs of a-t, v-t and s-t for an object released from rest in a uniform gravitation filed where there is air resistance.

Answer 21

Pg 13 of notes

Question 22

Why is the time of flight upwards for a bouncing ball less than its time of light downwards?

Answer 22

On its way up, the drag force acts as an extra retarding force to the object’s motion. Hence, velocity reduces to zero faster as the drag force is acting in the same direction as the object’s weight. Resultant force = Fd+mg = ma(up), a(up) = g+Fd/m, a(up) more than g. On the object’s way down after reaching maximum displacement, the drag force acts is in the opposite direction to the weight of the ball. Velocity increases at a slower rate. Resultant force = ma(down), a(down) = g-Fd/m, a(up) less than g. Hence, a(up) is greater than a(down). During the flight upwards, a shorter time is needed to cover the same distance. Thus, the time of flight upwards for a bouncing ball less than its time of light downwards.

Question 23

What is projectile motion?

Answer 23

It is a curved path in a vertical plane that involves motion in the x and y-directions simultaneously and the vertical and horizontal motions of a projectile are independent of each other.

Question 24

What can you say about the time taken for an object to travel in the x and y-directions?

Answer 24

They are the same.

Question 25

What are the general steps to analyse projectile motion?

Answer 25

(i) Resolve the displacement, velocity and acceleration vectors into their horizontal and vertical components (Sx, Sy, Ux, Uy, Vx, Vy, Ay, Ax is always 0) (ii) Apply the kinematic equations in each direction

Question 26

Name the 3 key characteristics of projectile motion.

Answer 26

(i) The path of the projectile is symmetrical about the vertical thought the highest point. Time taken to reach the highest point is equal to the time taken to go from the highest point to y = 0. (ii) Assumption that there is no air resistance hence horizontal acceleration is 0 and vertical acceleration a(y) = g directed downwards (iii) The path of a projectile is always a parabola.

Question 27

Define trajectory.

Answer 27

The path described by an object (parabolic for projectile motion) where the direction of velocity at each point is a tangent to the trajectory at that point. Note: Initial velocity will be stated and the direction of projection follows.

Question 28

Define range.

Answer 28

Horizontal displacement of the plane between the point of projection and the point of impact. It can be calculated by multiplying time to the horizontal velocity component.

Question 29

Define angle of projection.

Answer 29

The angle between the direction of projection and the horizontal plane through the point of projection.

Question 30

What effects does air resistance have on the projectile motion? Explain.

Answer 30

(i) The trajectory is no longer symmetrical because the projectile experiences air resistance in the direction opposite to its velocity. Hence, air resistance has both a horizontal and vertical component. Explanation same as (iii). (ii) A lower maximum height is reached because the vertical component of air resistance acts downward, meaning that Vy reduces to zero after moving a shorter vertical displacement. (iii) The range is shorter. The horizontal component of velocity decreases throughout the motion due to the horizontal component of the air resistance. Hence, the horizontal displacement of the projectile on its way down is also shorter than on the way up.

Question 31

Describe the v-t graph of a parachutist before the parachute is opened at terminal velocity.

Answer 31

Region A shows a line with constant gradient (i.e. acceleration of free fall g). Only gravitational force is significant as the speed is nearly 0 (air resistance is very small). In region B, the speed is increasing at a decreasing rate (i.e. acceleration is decreasing). Air resistance increases as speed increases, hence resultant downward force decreases. Region C shows the speed has reached a constant peak value (zero gradient and acceleration). Air resistance equals gravitational force (zero resultant force).

Question 32

When a ball leaves the thrower's hand, explain why it is possible for the ball to have a downward acceleration of approximately 20ms^-2.

Answer 32

After the ball leaves the thrower’s hand and moves upwards, it experiences both the force of gravity (its weight, mg) and drag force (kv) in the downward direction. When the ball just leaves the thrower’s hand, it has a high speed (v). Hence, the downward drag force is large. Resultant (downward) force on the ball, mg + kV = ma. Acceleration of the ball, a = g + kv/m > 9.81 m s-2Downward acceleration of the ball is larger than g and is largest when the ball just leaves the hand. Hence, the downward acceleration of 20 m s–2 is possible.

Question 33

When does an object have an acceleration of g in the presence of air resistance?

Answer 33

At the point when the ball is momentarily at rest. It does not experience any drag force (since v = 0) and its resultant force is just its weight. Hence, its acceleration at this point is equal to g.

Question 34

What is the requirement for the angle of projection?

Answer 34

It must be less than 90 degrees.