Question

Kinetic energy KE=1/2mv^2 if mass is increased by 4% And velocity reduced by 3% determine the approximate change in K.E... Apparently i need to use binomial theorem (a=x)^n but how do i do it?

Answer 1

Its really simple.

New mass = 1+4/100
i.e\[Mass_n = 1 + \frac{ 4 }{ 100 }\]

Answer 2

For velocity subtract

\[Velocity_n = 1-0.3 = 0.97\]

Answer 3

Square .97 now

Answer 4

Tell me you get .

Answer 5

For distinction I need to use binomial.... Thank you. I think i have done something like that.

Answer 6

The Binomial theorem example:
(x+y)^2 = x^2 + y^2 + 2xy

Answer 7

How can we possible apply the Binomial theorem here..??

Answer 8

I don't know. Will have to check with my teacher... Thank you for your help.

Answer 9

If I call with \(v_1,m_1\) the new values of the speed and mass, and with \(v_0,m_0\) the corresponding initial value, then I can write this:

\[\large \begin{gathered}
{m_1} = 1.04{m_0} \hfill \\
{v_1} = 0.97{v_0} \hfill \\
\hfill \\
\frac{{\frac{1}{2}{m_0}v_0^2 - \frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_0}v_0^2}} = \frac{{{m_0}v_0^2 - 1.04{m_0} \cdot {{\left( {0.97} \right)}^2}v_0^2}}{{{m_0}v_0^2}} = \hfill \\
\hfill \\
= \frac{{{m_0}v_0^2\left\{ {1 - 1.01 \cdot {{\left( {0.97} \right)}^2}} \right\}}}{{{m_0}v_0^2}} = ...? \hfill \\
\end{gathered} \]

Answer 10

values*

Answer 11

oops.. I have made a typo:
\[\large \begin{gathered}
\frac{{\frac{1}{2}{m_0}v_0^2 - \frac{1}{2}{m_1}v_1^2}}{{\frac{1}{2}{m_0}v_0^2}} = \frac{{{m_0}v_0^2 - 1.04{m_0} \cdot {{\left( {0.97} \right)}^2}v_0^2}}{{{m_0}v_0^2}} = \hfill \\
\hfill \\
= \frac{{{m_0}v_0^2\left\{ {1 - 1.04 \cdot {{\left( {0.97} \right)}^2}} \right\}}}{{{m_0}v_0^2}} = ...? \hfill \\
\end{gathered} \]

Answer 12

i see what your doing
this is like finding ΔΕ/Ε0 correct ?

Answer 13

Thank you.
.

Answer 14

that's right!! @alekos