Impulsive tension problem involving two particles

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Homework Statement


Particles A and B, each of mass m, are connected together by means of a light, inextensible string of length 2a. They are at rest a distance, a apart on a smooth horizontal plane. A is then projected with speed u along the plane at right angles to AB. Find the velocities of the particles immediately after the string becomes taut. I'm sorry I can't make any attempt at a solution as I don't quite understand what the question is asking of me. Can anyone shed some light on this and perhaps give some explanations along with your solution? Thanks in advance again! :D (Sorry for filling the forum with so many impulsive tension questions!) :redface:

The answers are as follows:
Particle A: u√7/4 AT 49.1 degrees to the string
Particle B: u√3/4 in the direction of the string

Homework Equations





The Attempt at a Solution

 
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hi jiayingsim123! :smile:

i don't understand why you're having such difficulty with these impulse questions :confused:

the general plan is to divide them into three parts … before during and after the "collision"

before and after, conservation of energy applies

during, you need to use the impulse equations your professor taught you (or conservation of momentum will usually work just as well)
jiayingsim123 said:
Particles A and B, each of mass m, are connected together by means of a light, inextensible string of length 2a. They are at rest a distance, a apart on a smooth horizontal plane. A is then projected with speed u along the plane at right angles to AB.

Find the velocities of the particles immediately after the string becomes taut. I'm sorry I can't make any attempt at a solution as I don't quite understand what the question is asking of me. Can anyone shed some light on this …

the mass moves at constant speed u until it is at a point C, distance 2a from A

(in this case, "before" and "after" aren't a problem …*everything is constant :wink:)

use trig to find the angle

at C, the string suddenly becomes tight, and there's a "collision"

so use the impulse equations …

what do you get? :smile:
 
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http://img571.imageshack.us/img571/5838/img1246xk.jpg

Hi there tiny-tim, here is my illustration of how I interpreted the question. I find it a tad strange that both the particles have different final speeds, I'm given the knowledge that if the string is already taut, both particles will move together at the same speed? Could you help clear that up for me? And thanks for teaching me that method, my teacher didn't really teach this chapter all that well and is going to proceed to another chapter after summer. Thanks in advance again! :)
 
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hi jiayingsim123! :smile:

that looks ok! :smile:

(and the dotted line should of course be marked "a")
jiayingsim123 said:
I find it a tad strange that both the particles have different final speeds, I'm given the knowledge that if the string is already taut, both particles will move together at the same speed?

no, the only constraint is that the distance between them must be constant …

i haven't worked it out o:), but i expect that, after the "collision", they'll circle round each other (with constant angular velocity), while their centre of mass moves in a straight line (in which direction? :wink:)
 
Hi there tiny-tim,

But I thought the distance between the two particles will increase? From a to 2a? Distance a is the initial distance (which is when the string is still not taut) and distance 2a is the final distance (which is when the string is already taut and impulse is already created in the string)? Thanks! :D

If that is the case, is the angle between A'B' (when string is already taut and the distance between the two particles is 2a) and AB (original situation in which the string is still slack) merely 60 degrees? (cos-1(0.5))

:confused: