# Another Mechanics Problem Using Vectors

1. May 22, 2013

1. The problem statement, all variables and given/known data
the unit vectors i and j are parallel to the cordinate axey Ox and Oy. Two particles A and B each have a mass of 0.4kg, they both leave the origin O at the same time, and move in the x-y plane. The velocity vectors at t time for A and B are $(4t\hat{i} + 6t\hat{j})m/s$ and $(3t^2\hat{i}-2\hat{j})m/s$ respectively.

A: calculate in Joules the kinetic energy of B when t=2
B: Find the acceleration vector of A and prove the force acting on A is constant
C: Calcualte to the nearest 0.1N the magntiude of F and find to the nearest degree the angly between Oy and the line of action of F.
D:Find the distance between A and B when t=2.

I am unsure on all of them to be honest, I am not sure if I am using the vectors correctly and especially my integration as I cant see where any integration constants are.

2. Relevant equations

KE=0.5mv^2
F=ma

3. The attempt at a solution
Part A:
First I find the magnitude of the velocity of B
$$\dot{r_b}= 3(2)^2\hat{i} - 2\hat{j} \\ \dot{r_b} = 12\hat{i} - 2\hat{j} \\ |\dot{r_b}|= \sqrt{12^2 + 2^2} = 12.17m/s$$
and then use that in the KE formula
$$KE=\frac{1}{2}mv^2 \\ KE=\frac{1}{2}0.4(12.17)^2 \\ KE=0.2 \times 148.11 = 29.62J \\$$

Part B:
I differentiate to get the acceleration vector
$$\ddot{r_a}=4\hat{i}+6\hat{j}$$
and as the I and J components are not functions of time that means the force must be constant as the acceleration is constant. Is that correct?

Part C:
I use newtons second law to find F
$$F=0.4(4\hat{i}+6\hat{j}) \\ F=1.6\hat{i}+2.4\hat{j} \\ |F|=\sqrt{1.6^2+2.4^2}=2.9N$$
And then to get the angle i did..
$$arctan(\frac{2.4}{1.6})=56.31 \\ 90-56.31=34°$$
to the nearest degree, I took the 56.31 from 90 as the question wants the angle between F and the y-axis, i hope I done it correct.

Part D:
To find the distance I first found both of the position vectors by integrating.
$$r_a=∫(4t\hat{i}+6t\hat{j}) \ dt = 2t^2\hat{i}+3t^2\hat{j} \\ r_b=∫(3t^2\hat{i}+2\hat{j}) \ dt = t^3\hat{i}+2t\hat{j} \\$$
And it asks when t is 2 so then i found the position vectors for both when t=2
$$r_a=2(2)^2\hat{i}+3(2)^2\hat{j} = 8\hat{i}+12\hat{j} \\ r_b=(2)^3\hat{i}+2(2)\hat{j} = 8\hat{i}+4\hat{j}$$
And then taking b from a
$$(8\hat{i}+12\hat{j})-(8\hat{i}+4\hat{j})=8\hat{j}$$
and then the magnitude
$$\sqrt{8^2}=8m$$

Any help is appreciated. :)

Last edited: May 22, 2013
2. May 22, 2013

### BruceW

your answers to parts A,B,C all look correct. But in part D, you use
$$r_b=∫(3t^2\hat{i}+2\hat{j}) \ dt = t^3\hat{i}+2t\hat{j} \\$$
But I don't think this is right, because the velocity vector that was given is:
$$(3t^2\hat{i}-2\hat{j})m/s$$
Maybe you accidentally forgot the minus sign?
p.s. excellent name :)

3. May 23, 2013

Ah thanks for point that out!

So should it be...

$$r_b=∫(3t^2\hat{i}-2\hat{j}) \ dt = t^3\hat{i}-2t\hat{j} \\$$
and then taking the difference and magnitude then becomes..
$$(8\hat{i}+12\hat{j})-(8\hat{i}-4\hat{j})=16\hat{j} \\ \sqrt{16^2}=16m$$

4. May 23, 2013

### BruceW

yep. looks good, nice work! pretty neat too. it makes it easy to read, so thanks!

5. May 23, 2013