Calculating KE Gain with Velocity Vectors: Explained and Simplified

[phospho]

gain in KE = 0.5mv^2 - 0.5mu^2

where v and u is velocity

say we had an object moving with velocity (i + j), and after 5m it's velocity was (2i + j), would the gain in KE be:

$0.5m(2^2 + 1^2)^2 - 0.5m(1^2+1^2)^2$

or

$0.5(2i + j)^2m - 0.5(i+j)^2m$

Just confused what to do when vectors come into play

thanks

 

[HallsofIvy]

gain in KE = 0.5mv^2 - 0.5mu^2

where v and u is velocity

Technically, u and v are "speed" not "velocity". That's an important distinction when you are talking about vectors: "velocity" is a vector, "speed" is the magnitude of that vector.

say we had an object moving with velocity (i + j), and after 5m it's velocity was (2i + j), would the gain in KE be:

$0.5m(2^2 + 1^2)^2 - 0.5m(1^2+1^2)^2$

No. The magitude of the vector ai+ bj is $\sqrt{a^2+ b^2}$ so the square of the magnitude is $a^2+ b^2$ You should have $0.5m(2^2+ 1^2)- 0.5m(1^2+ 1^2)= 0.5m(3)$

or

$0.5(2i + j)^2m - 0.5(i+j)^2m$

That's wrong because you can't square vectors like that.

Just confused what to do when vectors come into play

thanks

Neither of the formulas you give is correct. You have to work with the magnitude of the vectors. And, again, you have to distinguish between the vector velocity and the scalar speed.

 

[phospho]

Technically, u and v are "speed" not "velocity". That's an important distinction when you are talking about vectors: "velocity" is a vector, "speed" is the magnitude of that vector.


No. The magitude of the vector ai+ bj is $\sqrt{a^2+ b^2}$ so the square of the magnitude is $a^2+ b^2$ You should have $0.5m(2^2+ 1^2)- 0.5m(1^2+ 1^2)= 0.5m(3)$


That's wrong because you can't square vectors like that.


Neither of the formulas you give is correct. You have to work with the magnitude of the vectors. And, again, you have to distinguish between the vector velocity and the scalar speed.

Okay, so if KE is 1/2mv^2, where v is a velocity vector (this is what my book says) then why do I take the magnitude of the velocity, rather than the actual velocity vector?

Here is an actual example:

I don't particularly need help with the question but, if I was to use conservation of linear momentum on this problem:

My book says exactly this:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$
where a body of mass $m_1$ moving with a velocity $u_1$ collides with a body of mass $m_2$ moving with a velocity of $u_2$. $v_1$ and $v_2$ are the velocities of the bodies after the collision.

Now using this concept on this problem, it would be $mu + Mv = mp + Mq$, however if the velocity vector was in the form of $ai + bj$, would it be $m(ai + bj) + M(ci + dj) = m(ei + fj) + M(gi + hj)$ I assume. Then for energy, why do I use the magnitude of the velocity, rather than the actual velocity vector?

I hope I've made my confusion clear, thanks for helping.

 

[HallsofIvy]

Your book is using the "dot product" (also called "inner product"). If v= ai+ bj+ ck then v.v= a^2+ b^2+ c^2 which is the same as the magnitude of the vector, squared. In either case, v^2 or |v|^2 is a number not a vector.

 

[phospho]

Your book is using the "dot product" (also called "inner product"). If v= ai+ bj+ ck then v.v= a^2+ b^2+ c^2 which is the same as the magnitude of the vector, squared. In either case, v^2 or |v|^2 is a number not a vector.

I haven't come across that yet.

I still don't understand why I need to use the magnitude of the vector for energy, but can use the actual vector for momentum.