Mechanics

Question 5 only has parts a and b?

http://www.mrbartonmaths.com/resources/A%20Level%20Past%20Papers/Mechanics%201/2010%20-%20June/M1BAQAJune10QP.pdf

Could somebody possibly explain Question 6c) please

I'd appreciate any help :-)

Firstly, you know the acceleration of object A is 1.96m/s^2.

It is also known that object A will accelerate downwards, while object B is accelerating upwards.

It is given that object A accelerates for 2 seconds before the string breaks.

So using the equation:

v = u + at,

v = 0 + 1.98*2

v = 3.92m/s

That is the velocity of A after 2 seconds.

It then asks to find the velocity when it hits the surface.

To calculate this, we take initial velocity to be 3.92m/s and distance to be 4m as given. Acceleration is now given by 9.8m/s^2 since the string as broken and the only acceleration acting on the object now is gravity.

We thus use the equation:

v^2 = u^2 + 2as

v^2 = 3.92^2 + 2*9.8*4

v^2 = 93.779.68

v = 9.68m/s

I'll leave part iii) for you to work on now, but bear in mind that B has an initial velocity directed in the opposite direction to gravity. Have a go and feel free to post your working if you get stuck.