A touch of Mechanics
Two identical blocks A and B are connected by a light, inelastic string of length 2 l . Initially, block B lies on a smooth horizontal table with height greater than 2 l , and block A is held up in the air with the string stretched, with a pulley exactly halfway between A and B .
Block A is then given a sudden impulse and gets a velocity of V 0 in the vertically downward direction. Both blocks start to move, B strikes the pulley and stops, and A continues to move until it hits the vertical side of the table. For the sake of this problem, the acceleration due to gravity is neglected.
Which of the following is/are correct?
A.
The speed of block
B
when it hits the pulley is
V
0
8
3
.
B.
The speed of block
B
when it hits the pulley is
V
0
.
C.
The speed of block
A
when it hits the wall is
V
0
8
3
.
D.
The speed of block
A
when it hits the wall is
2
V
0
.
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1 solution
Work done by gravity? Is gravity even to be considered here (see that motion of both blocks are in directions perpendicular to the force of gravity as it is mentioned 'horizontal')? I got the block A one very easily by conservation of angular momentum about the pulley while for B I had some doubts in my equations, so it was a kind of luck by chance. Let me clarify through with this and then I'll post a solution.
P.S. For B, I was thinking to use the work energy theorem as
T ⋅ ℓ = 2 1 m v B 2 − 0
where T is the tension in the string. I had troubles calculating for tension (I know it isn't constant and is equal to the centripetal acceleration on A, but I wasn't able to find it.) If you know how to approach from here or have some other method, please let me know. :)
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No bro its not eq to centripetal acc only.rather if we choose pulley as the origin then we get mg-Tcos\(\alpha = md/dt(dr/dt)-\((d\alpha/dt)^2 r .Where r is the distance of A from Origin and α denotes angle with horizontal.Rather the motion here is confined to be in a horizontal plane.The term "Smooth Vertical Walls" is misleading though.In Horizontal Plane its quite easy to study the Motion where as it becomes cumbersome to solve the set of eq given by Polar Coordinates.
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hmm yes .understood my blunder .thanks
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@Avi Solanki – Solution-See one thing that's quite clear from the problem is that block B will strike first the pulley. A simple analogy is we observe the acceleration along horizontal of A and B when the string makes an angle α with the horizontal.
For A its T cos α / m and for B its T / m . So B strikes the pulley first, then A hits the wall. When B strikes let its velocity be v so A's velocity has 2 components. Radial velocity is v(as along the same string) and Tangential is u .
Then, COAM gives u = v ( o ) / 2 . Now conserving Energy gives v 2 + u 2 + v 2 = v ( o ) 2 . Put u and v = v ( o ) √ 3 / 8 . B comes to rest now so A's radial velocity diminishes suddenly. Now, A undergoes Uniform Circular Motion to hit the wall with its tangential velocity = v ( o ) / 2 .
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@Spandan Senapati – Yeah, I found this today itself. I'm laughing at myself now because after solving for A's velocity all that was needed to be done is cons. of energy. Nice solution, it would be better if you post a well written solution for this. Thanks. :)
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@Tapas Mazumdar – You mean handwritten solutions?Thanks for suggestions.I will surely try my best.
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@Spandan Senapati – Whatever is your choice man. You do know Latex right? If yes, then please post one in Latex only. :)
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@Tapas Mazumdar – try the problem mechanics and magentism .i insist
@Spandan Senapati – Probably I did something wrong, but what about work done by gravity?
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@Sumanth R Hegde – Its actually a bit misleading.The motion takes place in a horizontal plane only.
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@Spandan Senapati – Yea that's y angular momentum and energy conservation wud yield an answer...... But the figure i as though gravity is acting along direction of V 0 ...I Initially took gravity into account, ended up nowhere(angular momentum will also be not conserved and judging the vertical distance travelled by the block is tough) and then i saw that the options did not depend on g and neglected it to get the answer :p
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@Sumanth R Hegde – Ya when the gravitational force is taken into consideration then we need to solve the system of T = m d / d t ( d r / d t ) , − T / m + g c o s θ = d / d t ( d r / d t ) − w 2 r , w = − d θ / d t and m r 2 w = L ( c o n s t a n t ) .Its clearly difficult to comment on the motion of the two blocks. I once tried solving this but this proved to be quite difficult.So gravity case is too far from being considered a q worth asking in an exam(atleast not the mathematics).
@Spandan Senapati please post a solution.
@avi solanki Do mention that there is no influence of gravity
@Tapas Mazumdar did u neglect the work done by gravity ?