Mechanics

Question 1

Particle assumptions

Answer 1

• negligible dimensions, so forces will all act on a particle at the same point
• this means air resistance and rotational forces don’t act on the object (they can be ignored)

Question 2

Rod/beam (3)

Answer 2

Long rigid particle, only 1 dimension isn’t negligible
* Mass is concentrated along a line
* No thickness
* Rigid (does not bend)

Question 3

Thin

Answer 3

Negligible thickness

Question 4

Rigid

Answer 4

Doesn’t bend

Question 5

Smooth pulley

Answer 5

String is wwrapped around a disc (pulley)
* “Smooth” = no frictional force between the surfaces
* Tension in the string is the same on either side of the pulley

Question 6

Rough

Answer 6

Frictional force acts during movement

Question 7

Light (1/2)

Answer 7

Has no mass

A light string has the same tension throughout

Question 8

Inextensible string

Answer 8

Doesn’t stretch - acceleration is constant
String = thin

Question 9

Lamina

Answer 9

Flat, thin, 2D object
Mass is distributed across the surface

Question 10

Lamina example

Answer 10

Sheet of paper

Question 11

Wire

Answer 11

Thin, inextensible, 1D length of metal

Question 12

Bead (2)

Answer 12

Particle with a hole in it so it can move on a wire or string.
* Moves freely along a wire or string
* Tension is the same on either side of the bead

Question 13

Peg

Answer 13

Fixed support for another object
Dimensionless

Question 14

The velocity of a particle is given by v = 3i +5j m s−1. Find:
a) the speed of the particle
b) the angle the direction of motion of the particle makes with the unit vector i.

Answer 14

a) speed = √(3² + 3²) = 5.83m/s
b) acute angle made between the vector and the x-axis
tan(a) = 5/3 so a = 59°

Draw a diagram - the axes must meet at the starting location. This will show you the angle that needs to be found

Question 15

The acceleration of a motorbike is given by a = 3i − 4j m/s. Find the angle the direction of the acceleration
vector makes with the unit vector j.

Answer 15

tan(3/4) = 0.75
θ = 36.9° (to 3 s.f.)
Angle required = 180° – 36.9° = 143.1°

j acts in the +ve y direction so the angle will be obtuse

Draw a diagram - the axes must meet at the starting location. Using alternate angles will show you the angle that needs to be found

Question 16

The acceleration of a boat is given by a = −0.05i + 0.15j m/s. Find the angle the direction of the acceleration vector makes with the unit vector i.

Answer 16

tan θ (0.15/0.05) = 3
θ =71.6° (to 3 s.f.)
Angle required = 180° – 71.6° = 108° (to 3 s.f.)

Draw a diagram - the axes must meet at the starting location. This will show you the angle that needs to be found

Question 17

A plane flies from P to Q.
The displacement from P to Q is 100i + 80j m. Find the angle the vector PQ makes with the unit vector j.

Answer 17

tan(150/50) = 3
θ = 71.6° (to 3 s.f.)

Draw a diagram - the axes must meet at the starting location. Using alternate angles will show you the angle that needs to be found

Question 18

A man walks from A to B and then from B to C. His displacement from A to B is 6i + 4j m. His displacement from B to C is 5i − 12j m.
What is the magnitude of the displacement from A to C?

Answer 18

AC = AB + BC
AC = (6i + 5i
4j - 12j)
AC = (11i -8j)
AC = √(11² + (-8)²) = 13.6 km

magnitude not total distance

Question 19

A man walks from A to B and then from B to C. His displacement from A to B is 6i + 4j m. His displacement from B to C is 5i − 12j m.
What is the total distance the man has walked in getting from A to C ?

Answer 19

Total distance = |AB| + |BC|

|AB| = √(6² + 4²) = 7.21 km
|BC| = √(5² + -12²) = 13 km
Total distance = 7.21 + 13 = 20.21 km

total distance not magnitude

Question 20

velocity

Answer 20

The rate of change of displacement

Question 21

velocity of projection

Answer 21

The velocity with which the projectile is projected

Question 22

time of flight

Answer 22

The total time that an object is in
motion from the time it is projected to the time it hits the ground

Question 23

You throw a ball up in the air. It travels upwards, stops, then accelerated downwards until it hits the ground. In which direction is u/v positive?

Answer 23

Either direction - you can pick, but you must keep it the same.

Question 24

What acceleration would you use when modelling the path of a rock thrown upwards until it hits the ground?

Answer 24

a = -9.8m/s²

Question 25

How do you derive v = u + at?

Answer 25

* Sketch a velocity-time graph, with v and u as their corresponding velocities * The change in velocity = v - u * The acceleration = (v - u) / t * Rearrange to get v = u + at

Question 26

How do you derive s = (u + v) / 2 * t?

Answer 26

* Sketch a velocity-time graph, with v and u as their corresponding velocities (u isn't 0) * Displacement = area under graph = area of trapezium = (u + v) * t / 2 * Rearrange to get s = (u + v) / 2 * t

Question 27

How do you derive v² - u² = 2as?

Answer 27

* Rearrange v = u + at for t and substitute into s = (u + v) / 2 * t * t = (v - u) / a * s = (u + v) / 2 * ((v - u) / a) * s = (v² - u²) / 2a * Rearrange to get v² - u² = 2as

Question 28

How do you derive s = ut + 1/2 at²?

Answer 28

* Substitute v = u + at into s = (u + v) / 2 * t * s = (u + [u + at]) / 2 * t * s = (2u + at) / 2 * t * s = (u + 1/2at) * t * s = ut + 1/2at²

Question 29

How do you derive s = vt - 1/2 at²?

Answer 29

* Rearrange v = u + at for u and substitute into s = (u + v) / 2 * t * u = v - at * s = ([v - at] + v) / 2 * t * s = (2v - at) / 2 * t * s = (v - 1/2at) * t * s = vt - 1/2ut²

Question 30

How do you solve mechanics problems where an object is moving in one direction?

Answer 30

F = ma Expression for the resultant force Equate these two expressions

Question 31

Explain Newton's Third Law.

Answer 31

For every action there is an equal and opposite reaction. When two bodies A and B are in contact, if body A exerts a force on body B, then body B exerts a force on body A that is equal in magnitude and acts in the opposite direction.

Question 32

A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. How would you find the force exerted by mass A on mass B?

Answer 32

Find the normal reaction force of A. | "**by** A on B" = resolve for A

Question 33

A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. How would you find the force exerted on mass B by the scale-pan?

Answer 33

Find the force on the scale-pan by B. (The force exerted on B by the scale-pan has the same magnitude but acts in the opposite direction as this force - Newton's Third Law.) | "on mass B **by** the scale-pan" = resolve for the scale-pan

Question 34

A scale-pan is hanging by a string and has two boxes stacked on top of each other - A on top of B. Can the boxes A and B considered by themselves?

Answer 34

Yes for A but no for B, as the scale-pan is acting on B but not A

Question 35

A toy train travels along a straight track, leaving the start of the track at time t = 0. It then returns to the start of the track. The distance, s metres, from the start of the track at time t seconds is modelled by s = 4t² − t³, where 0 ≤ t ≤ 4. Explain this restriction on t.

Answer 35

Motion begins at t = 0 therefore t can't be less than 0. s is distance from the start of the track (or s is a distance so cannot be negative) so s is ≥ 0 4t² - t³ ≥ 0 t²(4 - t) ≥ 0 t² ≥ 0 for all values of t and (4 - t) is non-negative for t ≤ 4 therefore t²(4 - t) is non-negative when t ≤ 4 Hence 0 ≤ t ≤ 4 | Time can't be negative, although the question mentions t starts at 0s.

Question 36

How do you find the maximum/minimum velocity, given a formula for the velocity?

Answer 36

Differentiate the formula for the velocity (you must differentiate the thing you're finding the max/min of)

Question 37

How do you find the velocity, given a formula for the acceleration?

Answer 37

Integrate the formula for the acceleration

Question 38

When do you do the modulus of a definite integration?

Answer 38

When the curve goes under the x-axis (otherwise it would result in a negative area)

Question 39

A uniform rod of mass m is smoothly hinged at A against a wall. The rod is held in equilibrium by a horizontal force P at the other end of the rod. How would you find the magnitude and direction of the normal contact force at the hinge?

Answer 39

Draw a force triangle; P is the only force acting towards the wall and mg is the only force acting downwards. Therefore the normal contact force at the hinge must be equal in magnitude but act in the opposite direction to this resultant force. From the force triangle you can easily use tan_-1 to find the angle of the normal contact force to the horizontal/vertical.