Homework Problems

[Jimmy G]

Hey guys this is out to everyone doing the homework and having problems.
In this Topic we can discuss and help each other.
Just post on what questions yous are having problems with, how you tried to attempt it (if you want to ) and if anyone else can help you out if they know how to do it.

For example
I am stuck on 24/25 on the sheets and the last two questions in the past papers, the UFO and hot air balloon one.
If anyone has an queries on how to solve this or the formulas needed could you post a reply? Please and thank you!

[Admin]

James, here are a few pointers to help you:
1. Q24 and Q25 the force downslope is mg sin (angle of slope)
2. The UFO, upwards force is from the engines, downwards force is the weight but remember that g is only 9.8 on Earth.
3. The balloon, upwards force from the hot air, downwards forces are weight and tension (vertical component). If the balloon is stationary Fu = 0.

I didn't know if you were wanting me to post help or to leave it to your colleagues. If you are still struggling with these hints then let me know and I'll give you more help.

[S.McGrath]

why did u give us soo much homework over the holidays sir?

[Admin]

The more problems you do the better you become and since you are not in school for 2 days you have more time to do more problems and so become great at physics. Either that or I thought how will I ruin everyone's holiday! with homework of course.

[S.McGrath]

dont understand 21 or 22 so far

[Brian_c92]

sir am stuck on the questions on the sheet you gave us its the 2 questions after
the multiple choice

[Admin]

Steven & Brian,
I do not have the problems to hand, give me the information from the problem and what it is asking and I will be able to give you pointers.

[S.McGrath]

21.A rocket on a launch pad has two forces acting on it. (60000N up, 24500N down)
a)calculate the initial acceleration of the rocket
b)as the rocket rises what happens to the mass?
c)how does this affect the acceleration?

22. A 1kg mass is hung from a spring balance which is attached to the roof of a lift

In an upward journey the lift accelerates, moves at a constant velocity and decelerates to rest.
find the reading on the spring balance:
a) when the lift is stationary
b)when the lift is accelerating up at 3ms-2
c)when the lift is moving up at a constant velocity
d)when the lift is decelerating up at 3ms-2

[S.McGrath]

also Q23

duncan(mass 75kg) stands on bathroom scales inside a lift

on an upward journey the lift accelerates for 2s, moves at a constant velocity for 6s and decelerates to rest in a further 2s
wat does duncan read on the scales:
a)when is the lift stationary?
b)when the lift is accelerating up if the acceleration is 2ms-2
c)when the lift is moving at a constant velocity
d)when the lift is decelerating up if the deceleration is 2ms-2

[Brian_c92]

A lunar landing craft descends vertically towards the surface of the moon with a constant speed of 20m/p s-1. the craft and crew have a combined mass of 15000 Kg

Assume that the gravitational feild strength on the moon is 1.6N Kg -1


(a) During the first part of the descent the upward thrust of the rocket engine is 24 000 N
Show the results in the craft moving with a constant speed.

(b) The upward thrust of the engine is increased to 25 500N for the last 18 seconds of the descent.
(i) Calculate the deceleration of the craft during this time
(ii) What is the speed of the craft just before it lands ?
(iii) How far is the craft above the surface of the moon when the engine thrust is increased to 25 500N

[Brian_c92]

2(a) A hot air balloon, of total mass 500Kg, is held stationary by a single vertical rope.
(ii) When the rope is released, the balloon initially accelrates vertically upwards at 1.5m/s-2 find the magnitude of the bouyancy it is released
(iii) Calculate the tension in the rope before it is released

[S.McGrath]

i didnt get Q1 ^^^^^either

[Admin]

Sorry for the delay have been out all day,
Q21 a) Fu = T - W = 60000 - 24500 = 35500N; Remember W = mg so 24500 = 9.8 m so m = 24500 /9.8 = 2500 kg
Fu = ma therefore 35500 = 2500a; a = 35500/2500 = 14.2 ms-2.
b) As it rises it uses fuel so mass decreases.
c) I'll let you figure that one out look at the equations and the answer to part b.

Q22 Think lift problems Fu = T - W where T is the reading in the spring balance and Fu = ma for the 1 kg block.
a) Fu = ma = T - W; 0 = T - W therefore T = W = mg = 9.8N
b) Fu = ma = T - W; 3 x 1 = T - 9.8 therefore 3 + 9.8 = T; T = 12.8N
c) I'll let you figure that one out the answer to part a should help.
d) I'll let you figure that one out the answer to part b should help, just remember decceleration is negative acceleration.

Q23 Think lift problems Fu = T - W where T is the reading in the scales and Fu = ma for the 75 kg Duncan. Now the answers to 22 should help if you are still stuck let me know.

Q 1 The lunar lander
a Fu = Thrust - W; Thrust = 24000N, W = mg = 15000 x 1.6 = 24000N
Fu = 24000 - 24000 = 0 in other words ma = 0 so 15000a = 0; a = 0ms-2
b)i) Fu = ma = T - W; T - W = 25500 - 24000 = 1500 N; therefore 1500 = ma = 15000a: a = 0.1ms-2
b)ii) a = (v - u)/t: u = 20ms-1, v = ?, a = -0.1 ms-2, s = ?, t = 18s
v = u + at; v = 20 - 0.1 x 18 v = 18.2 ms-1
b)iii) s = ut + 1/2 ut2; s = 20 x 18 + (-0.05) x 18 x18; s = 360 - 16.2; s =343.8m

Q 2 The hot air balloon
a) ii) Fu = Thrust - Weight where Thrust in this case is bouyancy, Fu = 500 x 1.5 and Weight = 500 x 9.8. I'll let you do the maths.
a) iii) This is a trick question. You know the unbalaced force acting on the balloon when it is free and you also know that when it is tied to the rope that Fu is 0 due to the force from the rope pulling on the balloon. The tension in the rope cancels out the difference between Thrust - Weight.

[Brian_c92]

All hail king of physics
cheers sir

[S.McGrath]

einstein strikes again!

cheers sir