# The gravitational potential energy of an object is 81 J at a height of 7.34 m with mass of 1.13 kg. When it is released what is the momentum and impulse of the object when it strikes the ground?

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The object weighing 1.13 kg is placed at a height 7.34 m above the ground level. When the object is dropped it moves down due to the gravitational force of attraction of the Earth. As it moves down, the potential energy of 81 J is converted to kinetic energy.

When the object strikes the ground it has a potential energy of 81J. This is equal to (1/2)*m*v^2. Substituting the mass of the object

(1/2)*1.13*v^2 = 81

=> v = sqrt(162/1.13)

=> v = 11.97 m/s

The momentum of the object is given by m*v = 1.13*11.97 = 13.53 kg*m/s.

The impulse is the change in the momentum. Initially the momentum of the object is 0. The final momentum is 13.53. The impulse is therefore equal to 13.53 kg*m/s

If the gravitational potential energy of an object is 81 J at a height of 7.34 m with mass of 1.13 kg, what is the momentum and impulse of the object?

fist we need to plan the strategies:

Ep=mgh

Ep=81J

let

Ep=Ek ,as we know that we need V to find momentum and impulse

81=1/2mv^2

162=(1.13)V^2

143.36=V^2

11.97=V

hence,the velocity is 11.97m/s

Momentum is product of mass and velocity

then,

momentum=1.13(11.97)

=1352.61kg/ms^-1

Impulse is change in momentum,

then,

Ft=mv-mu ,m=1.13 u=0 v=11.97

=1.13(11.97)-1.13(0)

=1352.61kg/ms^-1

it is solved ..............................................................pop ye ye ye