• Got a question on how to use our new website? Check out our user guide here!

Length Dilation and Projectile Motion Questions (1 Viewer)

pokemonlv10

New Member
Joined
Aug 22, 2016
Messages
9
Gender
Male
HSC
2017
Hey so this is from a past paper and I just can't figure out how to do it.

19) The rest length of a train is 200m and the rest length of a railway platform is 160m. The train rushes past the platform so fast that , when observed in the patform's frame of reference, the train and the platform are the same length.

How fast is the train moving?

a) 0.6c -> is the answer
b) 0.75c
c) 0.8c
d) 1.25c

20) A ball is launched horizontally from a cliff with an intial velocity of u m/s. After two seconds, the ball's velocity is in the direction 45 degrees from the horizontal. What is the magnitude of the velocity in m/s at 2 seconds?
a) u
b) 1.5u
c) 19.6
d) 27.7 -> is the answer

I've tried drawing them out, rearranging formulas bt i'm just not getting the answer.
 

pikachu975

I love trials
Moderator
Joined
May 31, 2015
Messages
2,405
Location
NSW
Gender
Male
HSC
2017
For q19 just use the formula:
Lv = L0*sqrt(1-v^2 /c^2)
Sub Lv = 160 and then sub in the options for v and you find whichever one gets you L0 (200m) which is v = 0.6c.
 

pokemonlv10

New Member
Joined
Aug 22, 2016
Messages
9
Gender
Male
HSC
2017
Oh thanks you for question 19), i forgot that u can just use the answers from multiple choice and sub it in!
 

pikachu975

I love trials
Moderator
Joined
May 31, 2015
Messages
2,405
Location
NSW
Gender
Male
HSC
2017
For q20:

let V be the speed at 2 seconds
let down be the positive direction

Vy = Vcos45

Sub this into Vy = Uy + at, Uy = 0 because there is no vertical velocity at the start
Vcos45 = 9.8(2)
V = 27.7 (D)

So I guess the initial information of u m/s is a distractor
 

BenHowe

Active Member
Joined
Aug 20, 2015
Messages
346
Gender
Male
HSC
2016
Uni Grad
2020
It is launched horizontally, so initially you can consider it to be at it's turning point. Draw a diagram it helps visualise it
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top