M1 Question

A particle P moves in a straight line across a horizontal surface with retardation of magnitude 1.8ms^-2. The particle P passes through a point A with velocity 5ms^-1. It subsequently passes through the point B, where distance AB = 2.5m

Q: Find the velocities of P on the two occasions when it passes through B.

Reply 1

nota bene

SUVAT...

a=-1.8
s=2.5
u=5.0 to my understanding...

Then we have

v^2=u^2+2as

and you should have a quadratic.

(though I have trouble to see how it passes B twice! I'm not so sure the negative solution of the quadratic makes that much sense... never trust me on mechanics - I just answered this as noone else seemed to do it)

Reply 2

mikejones

nota bene

SUVAT...

a=-1.8
s=2.5
u=5.0 to my understanding...

Then we have

v^2=u^2+2as

and you should have a quadratic.

*cries*

Precisely what i did. Came out to 5(point something), yet the answer in the back of the bookj says +/- 4 :confused:

Ive never seen the word retardation (lol) used in this context so presumed i was reading the question all wrong.

meh, i'll just presume we're right and the answers sheet is wrong (unless anyone else wants to contribute...)

Reply 3

nota bene

The answer is

\pm 4

, remember retardation is negative, so you get

\sqrt{5^2-2\times1.8\times2.5}

Reply 4

mikejones

nota bene

The answer is

\pm 4

, remember retardation is negative, so you get

\sqrt{5^2-2\times1.8\times2.5}

I see. So retardation = deccelleration. Is there a fancy alternative to accelleration?

Also, could you explain what is actually going on in this scenario :redface:

Im struggling to visualise how it could pass the point B twice...

Thanks

Reply 5

nota bene

mikejones

I see. So retardation = deccelleration. Is there a fancy alternative to accelleration?

Uhm, I stumbled upon 'retardation' during the national Swedish physics competition and I was totally confused, but guessed what it was :p:

. I hope there is nothing similar for acceleration, hehe.

mikejones

Also, could you explain what is actually going on in this scenario :redface:

Im struggling to visualise how it could pass the point B twice...

Thanks

I said this in my edit in the first post - I don't think it makes much sense with the -4 solution, because it seems to me that it would mean it moves from B to A, which it doesn't (assuming it doesn't rebound or something silly).

Anyone else care to enlighten us?

Reply 6

mikejones

Im thinking perhaps its the exam boards way of saying, you get 2 answers but we cant be bothered to ask you to eliminate one :laugh:

Anyhow, the help was really appreciated nota, you have my respect :cool:

Reply 7

alex_hk90

mikejones

Also, could you explain what is actually going on in this scenario :redface:

Im struggling to visualise how it could pass the point B twice...

I'm only guessing but:
If it's decelerating when it passes point B the first time then at some point past point B it will start to travel in the opposite direction (its velocity becomes negative), and pass through B again (but going the other way, so back where it originally came from).

Reply 8

zrancis

I'd agree, It has a positive velocity passing through the first point, after this, the body slows, stops and reverses its direction, hence it passes through the same point with the same speed but in the opposite direction- like a ball attached to an elastic string