AQA Maths Mechanics 1B Jan 23rd 2013 Watch
This is how I understood the boat question, then used sine rule twice (couldn't remember cosine) to get the speed.
180 - tan^-1 (4/3) = 127
Sin127 / 4 = Sin(theta) / 3
(theta) = 37
180 - 37 - 127 = 16.2
x / Sin16.2 = 3 / Sin 37
x = 1.4 ms^-1
1ai) t = 40 seconds (3 marks)
1aii) deceleration = -0.2/0.2 ms-2 (3 marks)
1bi) acceleration = 0.4 ms-2 (3 marks)
1bii) velocity = 40 ms-1 (3 marks)
1c) average speed = 22.4 ms-1 (2 marks)
2a) resultant = 7i + 8j N (2 marks)
2b) magnitude = 10.6 N (2 marks)
2c) acceleration = 2.12 ms-2 (2 marks)
2d) angle = 48.8o (3 marks)
3a) Diagram with R, F and 3g in correct directions (1 marks)
3b) R = 22.5N (2 marks)
3c) Friction = 4.5N (2 marks)
3d) acceleration = 4.80 ms-2 (3 marks)
3e) no air resistance? (1 marks)
4a) R = 520N (3 marks)
4b) tension exerted on trailer = 1280N (3 marks)
4c) tension exerted on the tractor = 1280N (1 marks)
5) 2 possible speeds of B after collision: 1.25 ms-1 or 2.75 ms-1 (6 marks)
6a) show that alpha = 53.1o (2 marks)
6b) magnitude of resultant velocity of boat on return journey = 1.4 ms-1 (6 marks)
7a) expression for v at time t: v = (6i + 2.4j) + (-0.8i + 0.1j)(t) (2 marks)
7b) r = (6i + 2.4j)(t) + (0.5)(-0.8i + 0.1j)(t2) + 13.6i (3 marks)
7c) t = 12seconds, distance travelled = 45.6m (7 marks)
8a) horizontal component = 16 ms-1 (2 marks)
8b) V = 20.6 ms-1 (5 marks)
8c) alpha = 39.1o (3 marks)
I think this may be about right, any comments or changes you think are needed just ask
Posted from TSR Mobile
Resistive force = 800N
Mass = 2400Kg
Resistive force = R
Mass = 3500Kg
Acceleration = 0.2 ms-2
Driving force = 2500N
First question asked you to work out R which is 520N.
Second question asked you for the tension exerted on the trailer which is 1280N.
Third question asked to state the tension exerted on the tractor which should be equal to the tension exerted on the trailer which is 1280N.
'A tractor, of mass 3500kg, is used to tow a trailer, of mass 2400kg, across a horizontal field. The trailer is connected to the tractor by a horizontal tow bar. As they move, a constant resistive force of 800 newtons acts on the trailer and a constant resistance force of R newtons acts on the tractor. A forward driving force of 2500 newtons acts on the tractor. The trailer and tractor accelerate at 0.2ms-2.
a) Find R.
b) Find the magnitude of the force that the tow bar exerts on the trailer.
c) State the magnitude that the tow bar exerts on the tractor.'
For part (a):
Using F=ma and taking the mass of both the tractor and the trailer,
Driving force - resistive forces = mass x acceleration
2500 - 800 - R = (2400 + 3500) x 0.2
Rearranging for R,
R = 2500 - 800 - (2400 + 3500) x 0.2
R = 520N
For part (b):
Using F=ma but focussing on only the trailer this time,
Forward force - resistive forces = mass x acceleration
T - 800 = 2400 x 0.2
Rearranging for T,
T = 800 + 2400 x 0.2
T = 1280N
For part (c):
The tension exerted by the tow bar on the tractor and trailer must be equal therefore the force exerted by the tow bar on the tractor is also 1280N.