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A Level Maths Equations > Forces (2) > Flashcards

Forces (2) Flashcards

1
Q

A car is modelled as a particle travelling at a constant speed along a smooth horizontal surface. Explain two ways in which this model could be adjusted to be more realistic?

A

The road is unlikely to be smooth - as there’s friction between the road and the car

A constant speed is unrealistic - the model could account for how speed varies with time

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2
Q

A force of magnitude 7 N acts on a particle in the i-direction. Another force, of magnitude 4 N, acts on a particle with a direction of 30degrees. The resultant of these forces, R, has direction a

The magnitude of force R

A

Resolving the 4 N force gives:
4cos30degreesi + 4sin30degreesj N = 3.464…i + 2j N
So R = 7i + 3.464…i + 2j N =10.464…i + 2j N

Magnitude = ¬(10.464…^2 + 2^2) = 10.7 N

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3
Q

A force of magnitude 7 N acts on a particle in the i-direction. Another force, of magnitude 4 N, acts on a particle with a direction of 30degrees. The resultant of these forces, R, has direction a

The angle a

A

tan a = 2/10.464… = 10.8degrees

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4
Q

A sledge is held at rest on a smooth slope angled at 25degrees to the horizontal. The rope holding the sledge is at an angle of 20degrees above the slope. The normal reaction acting on the sledge due to contact with the surface is 80 N.

Find the tension, T, in the rope

A

Resolving horizontally:
Tcos(20degrees + 25degrees) - 80sin25degrees = 0
T = 80sin25degrees/cos45degrees
T = 47.8 N (3 s.f.)

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5
Q

A sledge is held at rest on a smooth slope angled at 25degrees to the horizontal. The rope holding the sledge is at an angle of 20degrees above the slope. The normal reaction acting on the sledge due to contact with the surface is 80 N.

The weight of the sledge

A

Resolving vertically:
W - Tsin45degrees - 80cos25degrees = 0
W = 106N

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6
Q

Two forces (xi + yj) N and (5i + j) N, act on a particle P of mass 2.5 kg. The resultant of the two forces is (8i - 3j) N.

Find the values of x and y

A

(8i - 3j) = (xi + yj) + (5i + j)
So, xi + yj = (8i - 3j) - (5i + j)
So x = 3 and y = -4

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7
Q

Two forces (xi + yj) N and (5i + j) N, act on a particle P of mass 2.5 kg. The resultant of the two forces is (8i - 3j) N.

The magnitude and direction of the acceleration of P

A

Using F_net = ma
8i - 3j = 2.5a
a = 3.2i - 1.2j ms^-2

Magnitude of a = ¬(3.2^2 + (-1.2)^2) = 3.42 ms^-2

tan a = -1.2/3.2 = -20.556…degrees

F_net acts down-right, so direction = 360degrees - 20.556…degrees = 339.4degrees

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8
Q

Two forces (xi + yj) N and (5i + j) N, act on a particle P of mass 2.5 kg. The resultant of the two forces is (8i - 3j) N.

Find the particle’s velocity vector, 5 seconds after it accelerates from rest

A

List variables: u = 0i + 0j, v = v, a = 3.2i - 1.2j, t = 5

v = u + at => v = 0i + 0j + 5(3.2i - 1.2j)
v = 16i - 6j ms^-1
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9
Q

A skydiver with mass 60kg falls vertically from rest, experiencing a constant air resistance force, R, as they fall. After 7 seconds, they have fallen 200m.

Find the magnitude of the air resistance on the skydiver to 3 s.f.

A 
s = 200, u = 0, a = a, t = 7
s = ut + (1/2)at^2 => 200 = (0 * 7) + (1/2) * a * 7
a = 200 * 2 / 49 = 8.163... ms^-2

F_net = ma => F_net = 60 * 8.163… = 489.79… N

Resolving forces, F_net = W - R = mg - R
489.79… = (60 * 9.8) - R
R = 98.2 N

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10
Q

A skydiver with mass 60kg falls vertically from rest, experiencing a constant air resistance force, R, as they fall. After 7 seconds, they have fallen 200m.

State any assumptions made in part A

A

The air resistance is not affected by the skydiver’s speed
The skydiver is a particle
Acceleration due to gravity is a constant 9.8ms^-2

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11
Q

A horizontal force of 8 N just stops a mass of 7 kg from sliding down a plane inclined at 15degrees to horizontal

Calculate the coefficient of friction between the mass and the plane to 2 d.p.

A

Resolving parallel
F_net = ma
8cos15degrees + F - (7g)sin15degrees = 7 * 0
F = (7g)sin15degrees - 8cos15degrees = 10.027… N

Resolving perpendicular
F_net = ma
R - 8sin15degrees - (7g)cos15degrees = 7 * 0
R = 68.333… N

Friction is limiting, so:
F = μR => 10.027… = μ * 68.333…
=> μ = 0.1467… = 0.15 (2 d.p.)

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12
Q

A horizontal force of 8 N just stops a mass of 7 kg from sliding down a plane inclined at 15degrees to horizontal

The 8 N force is now removed. Find how long the mass takes to slide a distance of 3 m down the line of the greatest slops

A 
Resolving perpendicular
R - (7g)cos15degrees = 7 * 0
R = 66.262... N
Friction is limiting, so:
F = μR => F = 0.1467... * 66.262... = 9.723... N

Resolving parallel
(7g)sin15degrees - F = 7a
7a = (7g)sin15degrees - 9.723…
a = (8.031…)/7 = 1.147…ms^-2

List variables: s = 3, u = 0, a = 1.147…, t = t
s = ut + (1/2)at^2
3 = (0 * t) + (1/2)1.147…t^2 => t^2 = 5.229…
t = 2.29 s

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13
Q

A 10 kg box is being held in equilibrium on a plane inclined at 30degrees to the horizontal by a force P acting parallel to the plane. The coefficient of friction between the box and the plane is 0.4

Find the range of possible values for the magnitude of the friction

A 
Resolving perpendicular
R - (10g)cos30degrees = 0
R = 84.87... N
F <= μR => F <= 0.4 * 84.87...
F <= 33.948... N
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14
Q

A 10 kg box is being held in equilibrium on a plane inclined at 30degrees to the horizontal by a force P acting parallel to the plane. The coefficient of friction between the box and the plane is 0.4

Hence find the range of possible values for the magnitude of P

A

If frictions acts up the plane
P + F - (10g)sin30degrees = 0
49 - P = F <= 33.948…
P <= 49 - 33.948… = P >= 15.05… N

If friction acts down the plane
P - F - (10g)sin30degrees = 0
P - 49 = F <= 33.948…
P <= 49 + 33.948… = P <= 82.94… N

So the possible range for P is 15.05… <= P <= 82.94… N

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15
Q

A car of mass 1500 kg is pulling a caravan of mass 500 kg. They experience resistance forces totalling 1000 N and 200 N respectively. The forward force generated by the car’s engine is 2500 N. The coupling between the two does not break

Find the acceleration of the car and caravan

A

Consider the car and caravan together

Resolving horizontally:
F_net = ma
2500 - (1000 + 200) = 2000a
a = 0.65 ms^-2

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16
Q

A car of mass 1500 kg is pulling a caravan of mass 500 kg. They experience resistance forces totalling 1000 N and 200 N respectively. The forward force generated by the car’s engine is 2500 N. The coupling between the two does not break

Find the tension in the coupling

A

Resolving horizontally for the caravan:
F_net = ma
T - 200 = 500 * 0.65
T = 525 N

17
Q

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley as shown. A has a mass of 7 kg and B has a mass of 3 kg. The particles are released from rest with the string taut, and A falls freely until it strikes the ground travelling at a speed of 5.9 ms^-1. A does not rebound after hitting the floor

Find the time taken for A to hit the ground

A

Resolving forces acting on A:
7g - T = 7a (1)
Resolving forces acting on B:
T - 3g = 3a => T = 3a + 3g (2)

Substituting (2) into (1)
7g - 3a - 3g = 7a => 4g = 10a => a = 3.92 ms^-2

List variables: u = 0, v = 5.9, a = 3.92, t = t
v = u + at
5.9 = 0 + 3.92t => t = 1.51 s (3 s.f.)

18
Q

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley as shown. A has a mass of 7 kg and B has a mass of 3 kg. The particles are released from rest with the string taut, and A falls freely until it strikes the ground travelling at a speed of 5.9 ms^-1. A does not rebound after hitting the floor

How far will B have travelled when A hits the ground?

A 
v^2 = u^2 + 2as
5.9^2 = 0^2 + (2 * 3.92 * s)
7.84s = 34.81
s = 4.44 m (3 s.f.)
19
Q

Two particles A and B are connected by a light inextensible string which passes over a smooth fixed pulley as shown. A has a mass of 7 kg and B has a mass of 3 kg. The particles are released from rest with the string taut, and A falls freely until it strikes the ground travelling at a speed of 5.9 ms^-1. A does not rebound after hitting the floor

Find the time (in s) from when A hits the ground until the string becomes taut again

A

When A hits the ground, speed of B = speed of A = 5.9 ms^-1.
B will continue to rise, momentarily stop and then fall freely under gravity. The string will be taut again when the displacement of B = 0

s = 0, u = 5.9, a = -9.8, t = t
s = ut + (1/2)at^2 = 5.9t - 4.9t^2

5.9t - 4.9t^2 = 0 => t(5.9 - 4.9t) = 0
t = 0 or t = 1.204….
So the stirng becomes taut again at t = 1.20 s (3 s.f.)

A Level Maths Equations (66 decks)

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