Exam revision Hard questions

[kenhung123]

OK I came into quite a lot of questions during revision so decided to make this thread.

A planet has a mass which is approximately 1/5 of that of the Earth, and its radius is about 1/2 of Earth. Let the gravity on the surface of Earth as g and the Earth's day as 24 hrs. Determine the gravity on the surface of the planet exerted by the planet itself in terms of g.

[kamil9876]

This is the acceleration that a body undergoes on the surface of a planet of radius R and mass M.

Suppose the earth had radius R and mass M, and the other planet has R' and M' and acceleration g'.




If I still remember physics...

[kenhung123]

Hmm...could you please explain why you equated the gravitational field of earth and planet?

[99.95]

A batsman hits a cricket ball from the ground. Air resistance can be ignored throughout the flight of the ball. The total time of flight of the ball is 4 secs and the ball travelled a horizontal distance of 120m.

Q27. find the height of the ball when it has travelled a horizontal distance of 80 m.

[Blakhitman]

A batsman hits a cricket ball from the ground. Air resistance can be ignored throughout the flight of the ball. The total time of flight of the ball is 4 secs and the ball travelled a horizontal distance of 120m.

Q27. find the height of the ball when it has travelled a horizontal distance of 80 m.

hmm...I'll try.

I'm assuming it's symmetrical, so I'd divide time by two and work out inital vertical velocity using a=-10 v=0 t=2 u=?. v=u+at to find initial vertical velocity, should get 20 ms^-1.

Then work out the horizontal velocity, v=120/4=30 so you can work out the time at 80m, t=80/30=2.667sec

Now we can use vertical components: u=20, t=2.667, a=-10 x=?, so we use x=ut+1/2at^2, and I got 17.8m. Is this correct?

[99.95]

A batsman hits a cricket ball from the ground. Air resistance can be ignored throughout the flight of the ball. The total time of flight of the ball is 4 secs and the ball travelled a horizontal distance of 120m.

Q27. find the height of the ball when it has travelled a horizontal distance of 80 m.

hmm...I'll try.

I'm assuming it's symmetrical, so I'd divide time by two and work out inital vertical velocity using a=-10 v=0 t=2 u=?. v=u+at to find initial vertical velocity, should get 20 ms^-1.

Then work out the horizontal velocity, v=120/4=30 so you can work out the time at 80m, t=80/30=2.667sec

Now we can use vertical components: u=20, t=2.667, a=-10 x=?, so we use x=ut+1/2at^2, and I got 40m. Is this correct?

nope the answ: 17.8m

[kamil9876]

Hmm...could you please explain why you equated the gravitational field of earth and planet?

I didn't, g' is the "g at that planet".

[shokstar]

Working horizontally:
u=?, t=4, x=120,
x=ut, 120=4u, u=30ms^-1
To find t at 80m:
u=30ms^-1, x=80, t=?
x=ut, 80=30t, t=2.67s

Now working vertically:
We need to find the initial vertical speed. To find that we work with the maximum point:
where V=o, t=2 (half of flight time), a=-10, u=?
v=u+at, 0=u-20, u=20ms^-1
Now finding x at t=2.67:
u=20ms^-1, t=2.67s, a=-10, x=?
x=ut+1/2at²
x=20x2.67+.5x-10x2.67²
x=17.7m
Sorry I don't know how to use that maxtext thing.

[99.95]

Working horizontally:
u=?, t=4, x=120,
x=ut, 120=4u, u=30ms^-1
To find t at 80m:
u=30ms^-1, x=80, t=?
x=ut, 80=30t, t=2.67s

thanks man!

Now working vertically:
We need to find the initial vertical speed. To find that we work with the maximum point:
where V=o, t=2 (half of flight time), a=-10, u=?
v=u+at, 0=u-20, u=20ms^-1
Now finding x at t=2.67:
u=20ms^-1, t=2.67s, a=-10, x=?
x=ut+1/2at²
x=20x2.67+.5x-10x2.67²
x=17.7m
Sorry I don't know how to use that maxtext thing.

[Blakhitman]

A batsman hits a cricket ball from the ground. Air resistance can be ignored throughout the flight of the ball. The total time of flight of the ball is 4 secs and the ball travelled a horizontal distance of 120m.

Q27. find the height of the ball when it has travelled a horizontal distance of 80 m.

hmm...I'll try.

I'm assuming it's symmetrical, so I'd divide time by two and work out inital vertical velocity using a=-10 v=0 t=2 u=?. v=u+at to find initial vertical velocity, should get 20 ms^-1.

Then work out the horizontal velocity, v=120/4=30 so you can work out the time at 80m, t=80/30=2.667sec

Now we can use vertical components: u=20, t=2.667, a=-10 x=?, so we use x=ut+1/2at^2, and I got 17.8m. Is this correct?

nope the answ: 17.8m

lol I did it correct, just didn't plug numbers in calculator correctly.

I just did it again, chuck in the numbers yourself.

[kenhung123]

I get it not you subbed g to replay GM/R^2.

That was smart. Thanks

[kenhung123]

Just as a general question: when solving electricity questions, do I necessarily need to state why I used a particular value e.g. I=V/R=0.1A but since they are in series therefore the LED is also 0.1A

[Blakhitman]

I just initiate by saying therfore

[Greggler]

can someone show me an example of a banked track question involving friction??

It was mentnioed in another thread.

I've only ever seen friction involved with incline planes, towing, etc. but ive never seen it involved with circular motion.

[shokstar]

Question below from STAV 08. q10.