Question

what is the kinetic energy when 1/2 mass reduced and 1/2 speed reduced?

Answer 1

Given:

final mass = $\frac{1}{2}$×initial mass

final velocity = $\frac{1}{2}$×initial velocity

To find:

The kinetic energy

Solution:

We know that,

Kinetic Energy$=\frac{mv^2}{2}$

Let the initial mass be $m_0$ and initial velocity be $v_0$, then the kinetic energy so produced will be,

$K.E._0=\frac{m_0v_0^2}{2}$

Now the mass is reduced to half its initial mass and velocity is reduced to half its velocity. Let the final mass be $m_f$ and final velocity be $v_f$, then the final kinetic energy becomes

$K.E._f=\frac{m_fv_f^2}{2}$

Now,

$m_f$ = $\frac{1}{2}$$m_0$ and $v_f$ = $\frac{1}{2}$$v_0$

So, putting the values,

$K.E._f=\frac{(\frac{1}{2}m_0)(\frac{1}{2}v_0)^2}{2}$

$K.E._f$ = $\frac{1}{8}$×$\frac{m_0v_0^2}{2}$

Hence, the final kinetic energy is reduced to half the initial kinetic energy.

Answer 2

Kinetic energy is One eight (1/8) when 1/2 mass reduced and 1/2 speed reduced

KE = (1/2) mv²

KE = Kinetic Energy

m = mass

v = velocity

Assume initial mass 2M and initial Velocity 2V

m = 2M

v = 2V

Hence Initial KE

KE = (1/2) 2M (2V)²

=> KE  = (1/2) 8MV²

1/2 mass reduced and 1/2 speed reduced

Hence Final mass = M  and Final velocity = V

Final KE =   (1/2)  M ( V)²

=>Final KE  = (1/2) MV²

=> Final KE  = (1/8)  (1/2) 8MV²

=> Final KE  = (1/8) Initial KE

Hence Kinetic energy is One eight (1/8) when 1/2 mass reduced and 1/2 speed reduced  

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