A body of mas 3 kg and a body of mass 10 kg are simultaneously dropped from a height of 14.9 m.Calculate their momenta,their potential energies and kinetic energies when they are 10 m above the ground. Please give the final answer also.ITS URGENT.

Let the oblect of mass 10 kg be A and the object of mass 3 kg be B

Formulae =

g = 9.8

P.E. = mgh 

P (momentum) = mv

K.E. = mv2/ 2

v2 of both A & B (it does not depend on mass) = u2 + 2as

OK so P.E. of A =

10 kg * 10 m * 9.8 m/s2

= 980 joule

P.E. of B =

3 kg * 9.8 m/s2 *10 m

= 294 joule

v2 = 0 + 2 * 9.8 m/s2 * (14.9 - 10) m

v2 = 96.04 m/s = 96 m/s (rounded)

v  = (root 96) m/s = 9.7979 m/s = 9.8 m/s (rounded)

K.E. of A =

10 kg  * 96 m/s * 1/2

= 480 joule

K.E. of B =

3 kg * 96 m/s * 1/2

= 144 joule

P of A =

10 kg * 96 m/s

= 960 kg m/s

P of B =

3 kg * 9.8 m/s

= 29.4 kg m/s

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So,

P.E. of A = 980 joule

P.E. of B = 294 joule

K.E. of A = 480 joule

K.E. of B = 144 joule

P of A = 960 kg m/s

P of B = 29.4 kg m/s

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