Question

A body of mass 3.0kg and a body B of mass 10 kg are dropped simultaneously from a height of 14.9m.Calculate their Momenta, their Potential energies and kinetic energies when they are 10m above the ground.

Answer 1

#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
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hope it helps!

Answer 2

Explanation:

the answer is here..........,.

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