Question

A body of mass 10kg is moving with a velocity (6i +8j) at a certain instant. it's K.E will be. A) 250J B) 500J C)100 J D) 1000J ...?

Answer 1

B) 500J

Explanation:

For 2D motion:
`v = v_x` i + `v_y` j

In our case:

`v = 6` i + `8` j `(ms^-1)`

Now `v^2 = v_x^2+v_y^2`
This can be seen using Pythagoras on a triangle with hypotenuse `v`, opposite `v_y` and adjacent `v_x` or by taking the dot product of the velocity vector with itself to find its length squared .

& `KE =1/2*mv^2` (by definition)

so substituting for `v^2` we get

-> `KE =1/2m(v_x^2+v_y^2)`

-> `KE =1/2*10*(6^2+8^2) J` (we are given mass is 10kg)

So plugging into a calculator (or using mental math):

`KE = 500 J`