Question

A small block slides with velocity 0.5 gr on the horizontal frictionless surface as shown in
the Figure. The block leaves the surface at point C. The angle o in the Figure is :
Vo
B
(A) cos(4/9)
(B) cos'(3/4)
(C) cos (1/2)
(D) none of the above​

Answer 1

Answer:

$(B) cos^{-1}(\frac{3}{4})$

Explanation:

1) Firstly, when small block leaves the surface at point C,then at that point Normal reaction by it must be zero.

Let the speed of block at point C be 'v'.

FBD of block is shown in figure attached .

Writing Force equation on small block :

$mgcos(\theta)-N=\frac{mv^2}{r}$

2) Then,

 Using Work Energy Theorem,

$mgr(1-cos(\theta))=\frac{mv^2}{2}-\frac{mv_0^2}{2}\\ \\=>mg(1-cos(\theta))=\frac{mv^2}{2r }-\frac{0.25mrg}{2r}\\ \\=>2mg(1-cos(\theta))=\frac{mv^2}{r}-0.25mg\\ \\=>2mg(1-cos(\theta))=mgcos(\theta)-0.25mg\\ \\=>cos(\theta)=\frac{3}{4}\\ \\=>\theta=cos^{-1}(\frac{3}{4})$

Hence, the angle O in the figure is as above.

Answer 2

SOLUTION

HEY BUDDY....

As we know if the body leaves the contact from any surface than the contact force between the surface and the body is zero .

Hence Normal force (N) is also zero...

Velocity At C be "V"

So ACC to FBD....(previous question)

mv²/r = mgcos θ - N. put N = 0

mv²/r = mgcos θ

v = √grcos θ.....(1)

Initial energy of the System....

1/2mu² ....

where u is the initial velocity of the body.

Final energy of the system....

1/2mv²+mg(r - rcos θ)....

As the body is moving on frictionless surface so...

FINAL ENERGY = INITIAL ENERGY

1/2mv² + mg(r - rcos θ) = 1/2mu²

(u = 1/2√gr)

(v = √grcos θ)

1/2m( v² - u² ) = -mg(r-rcos θ)

1/2( v² - u²) = -gr(1-cos θ)

v² - u² = -2gr(1-cos θ)

(√grcos θ)² - gr(1/4) = -2gr(1-cos θ)

grcos θ -(1/4)gr= -2gr(1-cos θ)

2gr - 1/4gr = 3grcos θ

9gr/4 = 3grcos θ

3/4 = cos θ

some thing wrong with the information you give

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