Question

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.

Answer 1

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  

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

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

b)

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