Mechanics 1 P42 2014; Discussion!

[Asad rehman]

what about second question?

 

[Mohamed1212]

I took paper 41 and it was very tough.

Here's the question I found the most difficult please help with it.

A car with mass 1100kg drives a distance 1760 m from A to B. Find the Increase in Kinetic Energy in terms of velocity V, and find the increase in P.E at distance x where0<=x<=1760. Then find K where KV^2=x.

 

[-V!p3R-]

Were you born on another planet? LOL! The GT in nov 2013 was less than 40 even though the paper was easy. GT will probably be below 36

Woah :O bro Chill! I was just making a random guess.
Tbh:- i believe in giving my best shot rather than expecting gt to be low So i don't know much about past gts

 

[Snowysangel]

Okay what was the answers to questions 5 to 7??

 

[Alfiko]

what about second question?

you'll get a quadratic equation

 

[thementor]

you'll get a quadratic equation

The answer was T =5 right?

 

[Ahsan Khan Niazi]

Are you kidding me..?? It was hell difficult paper from no where.

In my honest opinion, it was SOOOOO GOOOOODDDD
AL7AMDULILLAH *_*

 

[Oishee Asif]

Are you kidding me..?? It was hell difficult paper from no where.

;_; which ones did you think were hard?

 

[thementor]

;_; which ones did you think were hard?

Could you tell us the answers now?

 

[Oishee Asif]

Okay.
The tension answer thing with two strings was T2 = 11.9, T2 = 0.5. Checked a thousand times. The particle had weight too which you had to consider.

 

[Oishee Asif]

Answer of the integration question was 71.3m

 

[Oishee Asif]

Forgot what I wrote for number 7 completely.
For number 6,
a = -67.5, so deceleration = 67.5
R = 15.5, T = 17.something

 

[Oishee Asif]

1:
All I remember was last P was equal to 31.35kW

2. T = 5, forgot what the second part asked for

 

[thementor]

Forgot what I wrote for number 7 completely.
For number 6,
a = -67.5, so deceleration = 67.5
R = 15.5, T = 17.something

T=17.59 yessss

 

[Oishee Asif]

T=17.59 yessss

Yep

 

[thementor]

1:
All I remember was last P was equal to 31.35kW

2. T = 5, forgot what the second part asked for

P was 6150W
because at constant speed the forward force will be equal to the resistive force. So using P =FV, where F=410 and v=15

 

[thementor]

Yep

Man after hearing your answers I can relax.

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

 

[Oishee Asif]

P was 6150W
because at constant speed the forward force will be equal to the resistive force. So using P =FV, where F=410 and v=18

Are you sure F was 410?

 

[Oishee Asif]

Man after hearing your answers I can relax.

Q7 (I) TA = 0.25a +2.5, TB = 7.5-0.75a
(ii) a= 2 show
(iii) 1.2 ms^-1
(iv) -6 ms^-2 or simply the magnitude of the deceleration was 6

Ah yeah
Matched.

 

[thementor]

Are you sure F was 410?

Yupp.
See the mass of the car was 600kg and the acceleration was 1.4 and the Power given to you was 22.5 kW. and v=18