# Immediate Mechanics Help Required!!!

• Jun 6th 2008, 11:15 AM
AAKhan07
Immediate Mechanics Help Required!!!
http://www.meidistance.co.uk/npapers/m201ja.pdf
and see question 1(iv), the mark scheme answer is on page 7 but i need someone to explain how 24.6m/s was obtained.

Also, on page 8, how do you find 3.45 and 1.82 from those vectors on the top of the page? Thanks,
AAKhan07
• Jun 7th 2008, 01:38 AM
CaptainBlack
Quote:

Originally Posted by AAKhan07
http://www.meidistance.co.uk/npapers/m201ja.pdf
and see question 1(iv), the mark scheme answer is on page 7 but i need someone to explain how 24.6m/s was obtained.

Consevation of linear momentum. The initial momentum of the system is $300 \times20$ kg m/s the momentum of the component is $-20\times 44.5$ kg m/s. So the momentum of the remaining two parts is the initial momentum of the system plus $20\times 44.5$, so:

$280 \times v_{final} = 300 \times 20 + 20 \times 44.5$

RonL
• Jun 7th 2008, 01:43 AM
CaptainBlack
Quote:

Originally Posted by AAKhan07
Also, on page 8, how do you find 3.45 and 1.82 from those vectors on the top of the page? Thanks,
AAKhan07

Combine the vectors on the left into a single vector, then equate components on the right and left to get an equation for x and an equation for y. Solve these equations.

RonL