Question

A body of mass 3.0kg and a body B of mass 10 kg are dropped simultaneously from a height of 9m.Calculate their Momenta, their Potential energies and kinetic energies when they are 5m above the ground.

Answer 1

Explanation:

#POTENTIAL ENERGY

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PE = mgh

where m is mass of the body , g is acceleration due to gravity , h is the height above the ground.

• For 3Kg body :

PE at 10 m height above the ground is given by = 3×9.8×10 =294J

•For 10Kg body :

PE at 10 m height above the ground is given by = 10×9.8×10 =980J

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# MOMENTUM

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To find momentum of a body we need :

1) mass

2) velocity at that instant

we are provided with mass then we have to find velocities of both the bodies at the instant when they are at 10 m height above the ground after being dropped .

Finding Velocity :

Displacement = 14.9-10 = 4.9 m

Acceleration = g= 9.8m/s²

initial velocity(u) = 0 .

Velocity at 10m height = v (suppose)

time taken to cover the displacement = t (suppose)

using ,

S= ut +½×at²

we get ;

4.9 = ½×9.8×t²

=> t= 1 second

Then using ;

v= u + at

we get ;

v = 9.8(1) = 9.8 m/s

Now both the bodies are dropped simultaneously And WE HAVE NOT USED ANY MASS IN THIS PART .so both the bodies have velocity = 9.8 m/s at that instant.

•For 3Kg body :

Momentum = mv = 3×9.8

= 29.4 Kg m/s²

•For 10Kg body :

Momentum = 10×9.8

= 98 Kg m/s²

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# KINETIC ENERGY

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KE of a body is given by :

KE = ½× mv²

•For 3Kg body :

KE = ½×3×(9.8)²

= 144.06J

•For 10Kg body :

KE = ½×10×(9.8)²

= 480.2J

Answer 2

Answer:

When an object is dropped from a certain height above the ground, the only force acting on it is due to gravity, and its speed goes on increasing at a rate of 9.8m/s2 9.8 m/s2. The kinematic equations are applicable to a freely falling object.