(a) A particle of mass 4 kg moves under a force: F = 6t^2i - tj - 4k in Newtons.

Assuming that the particle is initially at the origin with velocity v = i + k in m/s, find its position after 1s.

 

(b) A ball of mass m travels with velocity 3 m/s and collides with a second ball with mass 2m at rest. After the collision the 2m ball moves with speed 1 m/s in a direction 60degrees to the original direction of the ball m.

(i) What is the final velocity of the ball m ?

(ii) What is the angle between the paths of the two balls after the collision ?

 

If you could answer and explain how to do either or even both of these it will possibly save me from failing an exam tommorow .

Hope you can help, Thanks !

What have you done so far towards answering these questions?

Honest answer is I dont know how to start on the question and Im just in desperate need of seeing a worked solution on it.

So for part (1)

 

You have two vectors, the initial velocity and the force acting.

 

Since you also have the mass, what can you calculate from the mass and the force?

 

What can you apply this result to to get a final distance travelled?

I think your suggesting integration. I dont understand integration but I can do differentiation. I can use F = ma if that would work ?

F=ma yes that's good.

 

So what can you calculate given mass and force?

 

Come on, you are nearly there.

 

 

Remember that for vectors you can split it down and just work with the components.

Ok i've tried this.

 

F = ma

F = ti - 6t^2j + 3k

v = j + 2k

mass = 4 kg

 

ti - 6t^2j + 3k = (4)a

 

ti - 6t^2j + 3k / 4 = a

 

V = u + at

 

j + 2k = 0 + (ti - 6t^2j + 3k / 4) 1

 

4(j + 2k) = ti - 6t^2j + 3k

4j + 8k = ti - 6t^2j + 3k

= ti - 6t^2j - 4j + 3k - 8k

= ti - 6t^2j - 4j - 5k

Edited April 29, 201213 yr by UrgentHelp

OK so you seem to have realised that you can calculate the acceleration.

 

Is the acceleration constant?

Is acceleration a scalar or a vector?

 

What about Klaynos' suggestion?

Edited April 29, 201213 yr by studiot

I dont understand how to split it down but since I have acceleration can I get that and sub it into s = ut + 1/2at^2 with the other values and solve for s ?

My questions are designed to highlight some point or other.

 

Acceleration is a vector in the same direction as the force.

Because it is in the same direction we can use the i, j k components of the force as the components of the acceleration vector if we divide each by the mass.

 

Using your original data I make the acceleration components as

 

 

[math]a = \frac{F}{m} = \frac{1}{4}\left( {6{t^2}i - tj - 4k} \right) = \frac{{6{t^2}}}{4}i - \frac{t}{4}j - k[/math]

 

 

You are introducing formulae for motion under constant acceleration.

 

 

I ask again is the acceleration constant?

 

 

I understand what your getting at. Whats the next step from that point. And no the acceleration is'nt constant. Btw i think you took down my force incorrectly. Mixed up the i and j ?

(a) A particle of mass 4 kg moves under a force: F = 6t^2i - tj - 4k in Newtons.

Assuming that the particle is initially at the origin with velocity v = i + k in m/s, find its position after 1s.

 

Compare this from post 1

 

Each of the i, j, k components are treated separately, as Klimatos said.

 

So the acceleration in the i direction is

 

 

[math]\frac{6}{4}{t^2}[/math]

 

etc

 

I assume that t is time so if you integrate this once with respect to time you will get a velocity along the i axis.

 

 

[math]{v_i} = \int_0^1 {\frac{6}{4}} {t^2}dt[/math]

 

Since we want to run for 1 second I have shown the integral limits 0 and 1

 

You must add the initial velocity in the i direction to this which according to your original post is +1

 

To get the distance moved along the integrate this velocity with respect to time.

This is the distance along the i axis and therefore the i coordinate.

 

You are asked to find the position after 1 second, not the distance travelled.

The position is given by the i, j and k coordinates in the same manner, although the integrations are much easier for j and k.

 

The distance travelled would be the length of the curve.

 

I have to go now, but others here seem keen to help.

Thanks for help !