JEE mechanics 1
Masses of 5 Kg and 3 Kg rest on two inclined plane each inclined at 3 0 ° to the horizontal and are connected by a string passing over the common vertex. After two seconds the mass of 5 Kg is removed. How far up (in meters) will the 3 Kg mass continue to move after removing the 5 Kg block?
g = 9 . 8 m / x 2
The answer is 0.6125.
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2 solutions
I did it the EXACT SAME WAY
I too did it the same way
Salaam, I have a confusion.... Why didn't the 3 kg block go down? In my opinion, as soon as the 5 kg weight is removed, the only force acting on the 3 kg block is the component of gravity along the wedge (neglecting friction). Kindly share the diagram and/or share a useful link if I'm overlooking some obvious concept.
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Hey, is the question like this? How far will the 3 Kg mass continue to move 2 seconds after removing Kg block?
The 3 kg block did not go down because the force and acceleration caused by the 5 kg block imparts some velocity in the 3 kg block which helps it in moving up, though momentarily because a force starts acting in the opposite direction causing deceleration =D
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@Hrishik Mukherjee – PLEASE DELETE YOUR COMMENT
THE FIRST LINE OF THE QUESTION SAYS THAT THE BLOCKS ARE AT REST. IT SHOULD BE REMOVED TO AVOID CONFUSION COZ LIKE THAT ANSWER IS 0
Hey ! Hey ! Hey ! My ans was also 0.6125 but i typed 0.6.... Sad {LAZY} :(
The gravitational force on both the blocks is gSin30=1/2*g. So the net force pulling 3 up is (5-3)*1/2*g. S o u p a c c 3 = m a s s m o v i n g d u e t o t h i s f o r c e f o r c e = 3 + 5 ( 5 − 3 ) ∗ 1 / 2 ∗ g = 8 1 g . S o u p V 3 , a f t e r 2 s e c = 8 1 g ∗ 2 = 4 1 g m / s . A f t e r 2 s e c , 3 w i l l m o v e u p a n d s t o p a f t e r m o v i n g u p a d i s t a n c e X u n d e r a g r a v i t a t i o n a l a c c . g S i n 3 0 = 2 1 g m / s 2 S o V 3 2 = 2 ∗ 2 1 g ∗ X . ⟹ X = ( 4 1 g ) 2 ∗ g 2 = 1 6 9 . 8 = 0 . 6 1 2 5 .
This question can easily be solved by drawing FBDs.
11
I have not marked the force acting on the blocks in direction perpendicular to the wedge because they are equal. Also M = 5 k g and m = 3 k g From the figure
M g s i n θ − T = M a .............(1)
and T − m g s i n θ = m a ...................(2)
Acceleration of both the blocks will be same because the string is in extensible.
Adding eq(1) and (2) and then after putting the value we get
a = 4 . 9 / 4
So after 2 seconds the velocity of the 3kg block will be u = 4 . 9 / 2
After the string is cut only a component of gravity acts in the direction acts in the direction parallel to the wedge. So the acceleration of 3kg block after cutting the string is g s i n ( 3 0 ) .
So S = 2 g s i n ( 3 0 ) u 2 Substituting the value we get S = 0 . 6 1 2 5 .