Two fixed frictionless inclined planes making an angle 30° and 60° with the vertical are shown in the figure. Two blocks  A and B are placed on the two planes. What is the relative vertical acceleration of A with respect to B

  • Option 1)

    4.9 ms -2 in vertical direction

  • Option 2)

    4.9 ms -2 in horizontal direction

  • Option 3)

    9.8 ms -2 in vertical direction

  • Option 4)

    Zero

 

Answers (1)

As we learnt in

When an Inclined Plane is given -

- wherein

R=mg\:cos\theta along normal to the incline

mg\:sin\theta =ma along the incline

a=g\:sin\theta

 

 Acceleration of the body down the smooth inclined plane a = g sin \theta

Vertical component of acceleration a is 

a_{(along\ vertical)}=(g sin\theta)sin \theta=g sin^{2}\theta

For block A

a_{A}=gsin^{2}60^{o}

For block B 

a_{B}=gsin^{2}30^{o}

\therefore    Relative vertical acceleration of A with respect to B

    =gsin^{2}60^{o}-gsin^{2}30^{o}

    =g\left[\left(\frac{\sqrt{3}}{2} \right )^{2}- \left(\frac1{2} \right )^{2}\right ]=\frac{g}{2}=4.9 ms^{-2}

Correct option is 1.

 


Option 1)

4.9 ms -2 in vertical direction

This is the correct option.

Option 2)

4.9 ms -2 in horizontal direction

This is an incorrect option.

Option 3)

9.8 ms -2 in vertical direction

This is an incorrect option.

Option 4)

Zero

This is an incorrect option.

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