a body A of mass 3 kg and a body B of mass 10 kg are dropped simultaneously from a height of 14.9 m .calculate:
(A) their momenta
(B) their potential energy
(C) their kinetic energy
when they are 10 m above the ground.

Velocity of A at the height of 10 m:v1=2gh1-h2    =2×9.8×14.9-10    =9.8 m/sVelocity of B at the height of 10 m:v2=2gh1-h2    =2×9.8×14.9-10    =9.8 m/s(a)Momentum of A:P1=m1v1     =3×9.8      =29.4 kg.m/sMomentum of B:P2=m2v2     =10×9.8      =98 kg.m/s(b)Potential energy of A:PEA=m1gh2         =3×9.8×10           =294 JPotential energy of B:PEB=m2gh2         =10×9.8×10           =980 J(c)Kinetic energy of A:KE1=12m1v12        =12×3×9.82          =144.1 JKinetic energy of B:KE2=12m2v22        =12×10×9.82          =480.2 J

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