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An explosion breaks an object into 2 pieces one is 1.5 times mass of the other. If 7,400 jules were released in the explosion, how much kenetic energy was aquired?   how much kenetic energy will each piece aquire

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jeew-m eNotes educator | Certified Educator

calendarEducator since 2012

write1,657 answers

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This is related to law of conservation of energy. It says that the energy is neither created nor destroyed.

So When the explosion takes place the energy released due to the explosion will be transferred to kinetic energy of the two pieces.

 

Let us say the mass of small peice is m and its velocity is v1.

Then the mass of bigger particle will be 1.5m and let us take the velocity of the bigger particle as v2.

 

Using momentum conservation;

Initial momentum = 0 (no movement before explosion)

Final momentum = `mv1+1.5mv2`

 

`0 = mv1+1.5mv2`

`v2 = -v1/1.5`

The (-) reveals that the parts were left apart from each other.

 

Using energy conservation;

Explosion energy `=` kinetic energy of two parts

                 `7400 = 1/2*m*(v1)^2+1/2*1.5m*(v2)^2`

                 `7400 = 1/2*m*(v1)^2+1/2*1.5m*((v1)/1.5)^2`

                ...

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jeew-m eNotes educator | Certified Educator

calendarEducator since 2012

write1,657 answers

starTop subjects are Math, Science, and Social Sciences

check Approved by eNotes Editorial


quantatanu | Student

As the object was at rest before explosion, so, 

Initial total momentum = 0  

Let the mass of one piece = m and speed v

     the mass of the other piece = m' and speed v'

Given, 

m' = 1.5 m   ------------------------(1)

From the momentum conservation, 

Initial momentum = Final momentum

=> 0 = m v + m' v'

=> m v = - m' v'

=> m v = - 1.5 m v'

=> v = - 1.5 v' ---------------------(2)

and the kinetic energy that the departed parts have acquired is actually acquired from the 7400 joules of the bomb. So energy conservation dectates:

 

7400 = (1/2) m v^2 + (1/2) m' v'^2

=> 2 * 7400 = m ( - 1.5 v' )^2 + (1.5 m) v'^2

=> 2 * 7400 = m*1.5*v'^2 [ 1.5  + 1]

=> 2*7400 = m*v'^2 * 1.5 * 2.5

=>  m*v'^2 = 2*7400/(1.5 * 2.5) --------------(3)

 

So, kinetic energy acquired by piece with mass m

= (1/2) m v^2

= (1/2) m (-1.5 v')^2

= (1/2)*(1.5)^2 m*v'^2

= (1/2)*(1.5)^2 *[2*7400/(1.5 * 2.5)]

= 4440 Joules

kinetic energy acquired by piece with mass m'=1.5 m

= (1/2)m' v'^2

= (1/2)*1.5*m*v'^2

=(1/2)*1.5* 2*7400/(1.5 * 2.5)

= 2960 Joules

 

To be confirmed we can add these kinetic energies to check that we get 7400 back,

4440 + 2960 = 7400

so the kinetic energy acquired by the two pieces are:

4440 Joules [piece with smaller mass]

&

2960 Joules [piece with bigger mass]

                  

 

quantatanu | Student

As the object was at rest before explosion, so, 

Initial total momentum = 0  

Let the mass of one piece = m and speed v

     the mass of the other piece = m' and speed v'

Given, 

m' = 1.5 m   ------------------------(1)

From the momentum conservation, 

Initial momentum = Final momentum

=> 0 = m v + m' v'

=> m v = - m' v'

=> m v = - 1.5 m v'

=> v = - 1.5 v' ---------------------(2)

and the kinetic energy that the departed parts have acquired is actually acquired from the 7400 joules of the bomb. So energy conservation dectates:

 

7400 = (1/2) m v^2 + (1/2) m' v'^2

=> 2 * 7400 = m ( - 1.5 v' )^2 + (1.5 m) v'^2

=> 2 * 7400 = m*1.5*v'^2 [ 1.5  + 1]

=> 2*7400 = m*v'^2 * 1.5 * 2.5

=>  m*v'^2 = 2*7400/(1.5 * 2.5) --------------(3)

 

So, kinetic energy acquired by piece with mass m

= (1/2) m v^2

= (1/2) m (-1.5 v')^2

= (1/2)*(1.5)^2 m*v'^2

= (1/2)*(1.5)^2 *[2*7400/(1.5 * 2.5)]

= 4440 Joules

kinetic energy acquired by piece with mass m'=1.5 m

= (1/2)m' v'^2

= (1/2)*1.5*m*v'^2

=(1/2)*1.5* 2*7400/(1.5 * 2.5)

= 2960 Joules

 

To be confirmed we can add these kinetic energies to check that we get 7400 back,

4440 + 2960 = 7400

so the kinetic energy acquired by the two pieces are:

4440 Joules [piece with smaller mass]

&

2960 Joules [piece with bigger mass]