# Thread: Immediate Mechanics Help Required!!!

1. ## 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

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

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

RonL

3. 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