Derive a formula for the fraction of kinetic energy lost

Asked by admin @ 05/08/2021 in Physics viewed by 254 People

Derive a formula for the fraction of the magnitude of kinetic energy lost. Express your answer in terms of the variables m and M.

b. Evaluate the fraction for m = 18.0 g and M = 380 g. Express your answer using three significant figures.

Answered by admin @ 05/08/2021

Given:

(a.)Derive a formula for the fraction of the magnitude of kinetic energy lost. Express your answer in terms of the variables m and M.

(b.) Evaluate the fraction for m = 18.0 g and M = 380 g. Express your answer using three significant figures

To find:

(a) $\frac{\Delta KE}{KE}$

(b) The fraction of the kinetic energy lost

Solution
:

• Consider a bullet of mass ‘m’ if fired in to a large block of mass ‘M’ suspended from some light wire

• Initially the bullet fired mean there is velocity of bullet that is v1 and block is at rest position

• When bullet hit the block the block swings through heigth ‘h’

Let us use the law of conservation of linear momentum

Initial momentum=final momentum

m1u1+m2u2=m1v1+m2v2 equation→ 1

In our Case m1 = m (mass of bullet)

m2 = M (mass of block)

u1 = initial velocity of bullet

u2 = initial velocity of block

v1 = final velocity of bullet

v2 = final velocity of block

Now initially our block is at rest mean u2=0

and final velocity of bullet and block is same so v1=v2=V

from equation 1
, We get

equation 1 → m1u1+m2u2=m1v1+m2v2

m1u1+m2(0)=m1V+m2V

m1u1+0=(m1+m2)V

m1u1=(m1+m2)V

V= $\frac{m1u1}{ (m1+m2)}$

Now the kinetic energy lost

ΔKE=KEi−KEf

ΔKE=0.5mv2−0.5(m+M) $\frac{m \times v}{(m+M)^2}$

ΔKE=0.5mv2− $\frac{m^2 }{ 2(m+M)v^2}$

The fraction of the magnitude of kientic energy lost

$\frac{\Delta KE}{KE}$ = $\frac{0.5mv^2-\frac{m^2 }{ 2(m+M)v^2}}{(0.5)(m)(v^2)}$

$\frac{\Delta KE}{KE}$ = $1-\frac{m}{(m+M)}$

m=18g

M=380g

$\frac{\Delta KE}{KE}$ = $1-\frac{m}{(m+M)}$

$\frac{\Delta KE}{KE}$ = $1-\frac{18}{(18+380)}$

$\frac{\Delta KE}{KE}$ =0.955

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