Inclined Plane

Classical Mechanics Level 3

A cubical block of mass m m is released from rest at a height h h on a frictionless surface of a movable wedge of mass M M , which in turn is placed on a horizontal frictionless surface as shown in the figure.

The magnitude of velocity of the triangular block when the smaller block reaches the bottom is

v = [ a m b g h cos ⁡ c θ ( m + M ) ( M + m sin ⁡ d θ ) ] e \large v = { \left [ \dfrac{a m^b g h \cos ^{ c }{ \theta } }{( m + M)( M + m\sin ^{ d }{ \theta } )} \right]}^{ e }

Enter your answer as a + b + c + d + e a + b + c + d + e .

D e t a i l s Details a n d and A s s u m p t i o n s Assumptions :

  • a a , b b , c c , d d are positive integers and e e is a positive rational number

  • Velocity is asked when the block reaches the bottom of the incline ( not the horizontal surface).

  • All surfaces of contacts are frictionless.

  • Size of block is negligible compared to size of the incline.

This is a part of my set Aniket's Mechanics Challenges .


The answer is 8.5.

Aniket Sanghi by Aniket Sanghi , 21, India

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4 solutions

Aniket Sanghi
Mar 24, 2016

Let the velocity of wedge be v 1 {v}_{1} towards right and v 2 {v}_{2} be the velocity of block relative to wedge along incline (downward) when the block has reached bottom.

  • Conserving Momentum :

M v 1 = m ( v 2 c o s θ − v 1 ) M {v}_{1} = m ({v}_{2}cos\theta - {v}_{1})

v 1 = m v 2 c o s θ m + M {v}_{1} = \frac{m {v}_{2}cos\theta}{m + M}

  • Conserving Energy :

m g h = 1 2 M v 1 2 + 1 2 m ( v 2 c o s θ − v 1 ) 2 + 1 2 m ( v 2 s i n θ ) 2 mgh = \frac{1}{2} M {v}_{1}^2 + \frac{1}{2} m ({v}_{2}cos\theta - {v}_{1})^2 + \frac{1}{2} m ({v}_{2}sin\theta)^2

now on solving you would get

v 1 = [ 2 m 2 g h cos ⁡ 2 θ ( m + M ) ( M + m sin ⁡ 2 θ ) ] . 5 \large {v}_{1} = { \left [ \dfrac{2 m^2 g h \cos ^{ 2 }{ \theta } }{( m + M)( M + m\sin ^{ 2 }{ \theta } )} \right]}^{ .5 }

26 Helpful 6 Interesting 9 Brilliant 3 Confused

Thanks!!!!

shivam mishra - 5 years, 2 months ago

Instead of velocity, you should use the word "Speed or magnitude of velocity" as it is more appropriate in this context as you haven't specified the direction of velocity.

Aditya Sky - 5 years, 2 months ago

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Oh! You were saying in the question , I have done the correction thanks! ....bye the way it was clear before also

Aniket Sanghi - 5 years, 2 months ago

I have specified.........One in earth frame towards right.......and the other relative velocity of block - down the incline

Aniket Sanghi - 5 years, 2 months ago

I did it using fbd in the frame of the wedge and found the time taken in terms of accln of wedge. Then I found the value of a in terms of m, M and theta. Then I did a x t = v, since initially the wedge was at rest

Rohit Nair - 5 years, 2 months ago

Can u post a solution taking velocities in earth frame

0 Helpful 1 Interesting 0 Brilliant 0 Confused
Shivam Mishra
Mar 24, 2016

@Aniket Sanghi can you please post a solution??

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Try it yourself again.....i am giving hint... Find the acceleration of the wedge and also the acceleration of the block along y (i.e. downward direction)....using the concept of pseudo force....or u can use conventional method also

Sujoy Ghosh - 5 years, 2 months ago

Both the acceleration will be constant......and we use the accleration along y dir. Of the block to find the time taken to come down and then multiply the time and acceleartion of wedge to find speed

Sujoy Ghosh - 5 years, 2 months ago

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Yes of course, one can do that. Doing so yields the expression of time taken by the block to reach the bottom of the swedge :- T = 2 h ( M + m ⋅ sin ⁡ 2 ( θ ) ) g ⋅ sin ⁡ 2 ( θ ) ( M + m ) \color{#3D99F6}{T\,=\,\sqrt \frac{2h(M+m \cdot \sin^{2}(\theta))}{g \cdot \sin^{2}(\theta)(M+m)}} . But, since we're suppose to figure out only the magnitude of velocity of the wedge by the time block reaches it's bottom, therefore, it is wise to use Principle of Conservation of Linear Momentum \color{#D61F06}{\text{Principle of Conservation of Linear Momentum}} and Work Energy Theorem \color{#D61F06}{\text{Work Energy Theorem}} for the said pupose as this takes less time.

Aditya Sky - 5 years, 2 months ago

Uploaded solution

Aniket Sanghi - 5 years, 2 months ago
Sujoy Ghosh
Mar 24, 2016

Please rectify the format of velocity of wedge.... The denominator is (M+m)(M+m (sin(ø))^2)... By mistake u have written m instead of M in 2nd term..

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Thanks! Done...

Aniket Sanghi - 5 years, 2 months ago

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