[herothing11]

A ball of mass 0.60kg rolls North at 4.0ms^-1.
It meets a slope which causes a force of 0.18N East.
This force lasts for 10s.
Calculate the final velocity of the ball, neglecting any friction.

I'm pretty stuck on this. The velocity northwards is 4 ms^-1 and that shouldn't change since the force is acting Eastwards right?

[Killjoy-]

Yes, so find the acceleration eastwards using F=ma and then consider the change in velocity (from 0) eastwards in the given time interval.

The velocity in the northern direction will be the same, so use that coupled with the answer above to find the final speed and direction of the ball.

[Stonebridge]

Tha velocity component due north stays the same.
What changes is the component due East. This is zero at the start but increases as a result of the force.
The final velocity will be the sum of the north and east components.

[herothing11]

Ah, right.
So the acceleration a = F/m
= 0.18/0.60
= 0.3 ms^-2

U = 0 V = V a = 0.3ms^-2 t = 10s

V=U+at
= 0.3 x 10
= 3ms^-1

Total Velocity = 4 + 3 = 7ms^-1?

[Killjoy-]

(Original post by herothing11)
Ah, right.
So the acceleration a = F/m
= 0.18/0.60
= 0.3 ms^-2

U = 0 V = V a = 0.3ms^-2 t = 10s

V=U+at
= 0.3 x 10
= 3ms^-1

Total Velocity = 4 + 3 = 7ms^-1?

All except part in bold is fine.

You will have to use pythagoras to find the resultant speed.

And then use tantheta=opp/ adj to find the direction it is moving. (And hence find the velocity.)

You could also give the answer as 4ms^-1 north and 3ms^-1 east.

[herothing11]

(Original post by Killjoy-)
All except part in bold is fine.

You will have to use pythagoras to find the resultant speed.

And then use tantheta=opp/ adj to find the direction it is moving. (And hence find the velocity.)

You could also give the answer as 4ms^-1 north and 3ms^-1 east.

Of course, excuse the stupidity
Root 4^2 + 3^2 = 5
Tan^-1(4/3)
=53 degrees from the 3N force

[Stonebridge]

Correct.

(Maybe 37 degs due East of North is the more common way of expressing the angle.)