Question

A body of mass 20g changes its linear momentum from 8kgm/s to 4kgm/s. calculate its work done.
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Answer 1

SOLUTION

GIVEN

A body of mass 20 g changes its linear momentum from 8 kgm/s to 4 kgm/s.

TO DETERMINE

The work done.

EVALUATION

Here it is given that a body of mass 20 g changes its linear momentum from 8 kgm/s to 4 kgm/s.

Mass of the body = m = 20 gm = 0.02 kg

Initial momentum $\sf{ = P_1 }$ = 8 kgm/s

Final momentum $\sf{ = P_2 }$ = 4 kgm/s

Thus initial kinetic energy

$\displaystyle \sf{ = \frac{1}{2m} {P_1}^{2} \: \: \: J}$

$\displaystyle \sf{ = \frac{1}{2 \times 0.02} \: \times \: {8}^{2} \: \: \: J}$

Final kinetic energy

$\displaystyle \sf{ = \frac{1}{2m} {P_2}^{2} \: \: \: J}$

$\displaystyle \sf{ = \frac{1}{2 \times 0.02} \times {4}^{2} \: \: \: J}$

Hence the required work done

$\displaystyle \sf{ = \bigg(\frac{1}{2m} {P_2}^{2} - \frac{1}{2m} {P_1}^{2} \bigg) \: \: \: J}$

$\displaystyle \sf{ = \frac{1}{2m} \bigg({P_2}^{2} - {P_1}^{2} \bigg) \: \: \: J}$

$\displaystyle \sf{ = \frac{1}{2 \times 0.02} \bigg({4}^{2} - {8}^{2} \bigg) \: \: \: J}$

$\displaystyle \sf{ = \frac{1}{ 0.04} \bigg(16 - 64\bigg) \: \: \: J}$

$\displaystyle \sf{ = - \frac{48}{0.04} \: \: \: J}$

$\displaystyle \sf{ = - \frac{4800}{4} \: \: \: J}$

$\displaystyle \sf{ = - 1200 \: \: \: J}$

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Answer 2

Given : A body of mass 20g changes its linear momentum from 8kgm/s to 4kgm/s.  

To Find :   work done.  

Solution:

mass = m = 20 g  = 20/1000  =  0.02  kg

momentum  =  mass * velocity

Initial momentum  =   8kgm/s

=> 0.02 * v₁  = 8

=>  v₁ = 400 m/s

Kinetic energy = (1/2)mv²

Initial  Kinetic energy = (1/2)(0.02)(400)²  = 1600 J

Final momentum  =   4 kgm/s

=> 0.02 * v₁  = 4

=>  v₂= 200 m/s

Final  Kinetic energy = (1/2)(0.02)(200)²  = 400 J

Work Done = Change in Kinetic energy

=  Final  Kinetic energy  - Initial  Kinetic energy

=  400 - 1600

= - 1200 J

= - 1.2 kJ

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