EFPBH's Physics 3/4 Question Thread

[EFPBH]

Marshall has a mass of 57 kg and he is riding his 3.0 kg skateboard down a 5.0 m long ramp that is inclined at an angle of 65 to the horizontal. Ingnore friction.

(i) calculate magnitude of the normal force acting on the Marshall and his skateboard.

why must we use Fn = mg sin

[PB]

Hey EFPHBH,
you aren't meant to use that formula?
That formula is for calculating the NET force acting on the skateboarder (down the ramp), causing him to accelerate down the plane.
The question is asking for you to find the NORMAL force acting on the skateboarder. In which case you should use Fnorm=mgcos(theta).
Maybe you might have thought that the little 'n' in front of the F meant normal instead of net. Be careful.

[RKTR]

During a game of Totem tennis, the ball of mass 150g is swinging freely in a horizontal circular path. the cord is 1.50m long and is at an angle of 60.0 to the vertical.

(i) calculate the radius of the ball's circular path.


explanation would be much appreciated

 

Draw a triangle . Radius= 1.50 x sin60

[PB]

^cord will be the hypotheneus of 1.50m. the angle the hypotheus makes with the verticle is 60 degrees.   so therefore, if you draw out that right angled triangle, you will find that the only side left unknown (aka the horizontal line) is the radius of the horizontal circle that the ball is drawing out in the air.

From there on, follow RKTR's method and use simple trig to solve it.

[rhinwarr]

Find the period of rotation, which is the time it takes for one lap.
T = 2pi r/v = (2pi x 450)/32 = 88.4s

3 minutes = 180s
180/88.4 = 2 laps

[Yacoubb]

A golf of mass 250g is hit horizontally from the top of a 40.0m high cliff with a speed of 25.0m/s. assuming a=9.8ms-2 due to gravity and ignore air friction.

calculate the distance that the ball travells from the base of the cliff ?

help is much appreciated as always   

I think I got the answer but could you tell me what the answer is so I show you the method? Thanks

[Yacoubb]

Answer: 71.5m

Yep I got that. Ok!

Using x = ut + 0.5at²

x = 40m
a = 9.8m/s/s
t = ?
u = 0m/s

40 = 0.5 * 9.8 * t²
t = 2.857142857142858

Then, you use x = ut to calculate the horizontal range.

x = ?
u = 25m/s
t = 2.857142857142858

x = 25 * 2.857142857142858 = 71.4m/s

[Stevensmay]

To calculate the velocity of the golf ball, we need to know it's velocity in the x direction and the y direction.
First recognize that . Since there will be no deceleration in the horizontal direction,

To calculate we use , where as per the question. Sub in and we find that

Now that we have our two velocities (that form two sides of a triangle), the overall velocity of the ball will be equal to the hypotenuse.



I hope this is correct because its been awhile.

[Thorium]

I think these type of questions are worth around 2 marks. So if u show abit of working, two-three steps, and the final ans in the appropriate form, it should be given full mark. Therefore I think ur solution should get full marks.

Btw, as "every second counts" in the exam, I would try to avoid writing unnecessary texts. For eg, instead of writing "therefore, try to use the symbol(the three dots) just to save a second in every question u do.

[PB]

I agree, get rid of the 'Therefore'
However, I personally don't think that this would get full marks. The examiner's report states that you need to write down the formula you are using.  sooo, you should really put m1v1+m2v2 = m3v3 or something like that! Always write down the formula you are using for every calculation you are doing (if a formula is being used that is), even in Chemistry.