Physics-
General
Easy

Question

A body of mass 2 blank k g slides down a curved track which is quadrant of a circle of radius 1 blank m e t r e. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is

  1. 4.43 blank m divided by s e c    
  2. 2 blank m divided by s e c    
  3. 0.5 blank m divided by s e c    
  4. 19.6 blank m divided by s e c    

The correct answer is: 4.43 blank m divided by s e c


    By conservation of energy, m g h equals fraction numerator 1 over denominator 2 end fraction m v to the power of 2 end exponent
    rightwards double arrow v equals square root of 2 g h end root equals square root of 2 cross times 9.8 cross times 1 end root equals square root of 19.6 end root equals 4.43 blank m divided by s

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    According to the graph the acceleration a varies linearly with the coordinate x. We may write a equals alpha x, where alpha is the slope of the graph.
    From the graph
    alpha equals fraction numerator 20 over denominator 8 end fraction m g subscript 0 end subscript equals 2.5 blank s to the power of negative 2 end exponent
    The force on the brick is in the positive x-direction and according to Newton’s second law, its magnitude is given by
    F equals fraction numerator a over denominator m end fraction equals fraction numerator alpha over denominator m end fraction x
    If x subscript f end subscript is the final coordinate, the work done by the force is
    W equals not stretchy integral from 0 to x subscript f end subscript of F blank d x equals fraction numerator a over denominator m end fraction not stretchy integral subscript 0 end subscript superscript x subscript f end subscript end superscript x blank d x
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    equals 8 blank J

    A 10 kg brick moves along an x-axis. Its acceleration as a function of its position is shown in figure. What is the net work performed on the brick by the force causing the acceleration as the brick moves from x equals 0 to x equals 8.0m?

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    According to the graph the acceleration a varies linearly with the coordinate x. We may write a equals alpha x, where alpha is the slope of the graph.
    From the graph
    alpha equals fraction numerator 20 over denominator 8 end fraction m g subscript 0 end subscript equals 2.5 blank s to the power of negative 2 end exponent
    The force on the brick is in the positive x-direction and according to Newton’s second law, its magnitude is given by
    F equals fraction numerator a over denominator m end fraction equals fraction numerator alpha over denominator m end fraction x
    If x subscript f end subscript is the final coordinate, the work done by the force is
    W equals not stretchy integral from 0 to x subscript f end subscript of F blank d x equals fraction numerator a over denominator m end fraction not stretchy integral subscript 0 end subscript superscript x subscript f end subscript end superscript x blank d x
    equals fraction numerator alpha over denominator 2 m end fraction x subscript f end subscript superscript 2 end superscript equals fraction numerator 2.5 over denominator 2 cross times 10 end fraction cross times open parentheses 8 close parentheses to the power of 2 end exponent
    equals 8 blank J
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    Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 blank m

    Work = Area under left parenthesis F minus d right parenthesis graph
    equals 8 plus 5 equals 13 blank J

    Force F on a particle moving in a straight line varies with distance d as shown in the figure. The work done on the particle during its displacement of 12 blank m

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    Work = Area under left parenthesis F minus d right parenthesis graph
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    The potential energy of a system is represented in the first figure. The force acting on the system will be represented by

    As slope of problem graph is positive and constant upto certain distance and then it becomes zero
    So from F equals fraction numerator negative d U over denominator d x end fraction, up to distance a,
    F equals constant (negative) and becomes zero suddenly

    The potential energy of a system is represented in the first figure. The force acting on the system will be represented by

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    As slope of problem graph is positive and constant upto certain distance and then it becomes zero
    So from F equals fraction numerator negative d U over denominator d x end fraction, up to distance a,
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    The work done by force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

    Work done W equals A r e a blank A B C E F D A
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    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 15 plus 10 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
    equals 125 plus 75 equals 200 J

    The work done by force acting on a body is as shown in the graph. The total work done in covering an initial distance of 20 m is

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    equals A r e aABCD +Area CEFD

    equals fraction numerator 1 over denominator 2 end fraction cross times open parentheses 15 plus 10 close parentheses cross times 10 plus fraction numerator 1 over denominator 2 end fraction cross times left parenthesis 10 plus 20 right parenthesis cross times 5
    equals 125 plus 75 equals 200 J
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    Two rectangular blocks A blankand B blankof masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 N m to the power of negative 1 end exponentand are placed on a frictionless horizontal surface. The block A blankwas given an initial velocity of 0.15 m s to the power of negative 1 end exponent in the direction shown in the figure. The maximum compression of the spring during the motion is

    As the block A moves with velocity with velocity 0.15 m s to the power of negative 1 end exponent, it compresses the spring Which pushes B towards right. A goes on compressing the spring till the velocity acquired by B becomes equal to the velocity of A, i.e. 0.15 m s to the power of negative 1 end exponent. Let this velocity be v. Now, spring is in a state of maximum compression. Let x be the maximum compression at this stage.

    According to the law of conservation of linear momentum, we get
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    Or v equals fraction numerator m subscript A end subscript u over denominator m subscript A end subscript plus m subscript B end subscript end fraction
    fraction numerator 2 cross times 0.15 over denominator 2 plus 3 end fraction equals 0.06 m s to the power of negative 1 end exponent
    According to the law of conservation of energy
    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
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    Two rectangular blocks A blankand B blankof masses 2kg and 3 kg respectively are connected by spring of spring constant 10.8 N m to the power of negative 1 end exponentand are placed on a frictionless horizontal surface. The block A blankwas given an initial velocity of 0.15 m s to the power of negative 1 end exponent in the direction shown in the figure. The maximum compression of the spring during the motion is

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    According to the law of conservation of linear momentum, we get
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    fraction numerator 1 over denominator 2 end fraction m subscript A end subscript u to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction open parentheses m subscript A end subscript plus m subscript B end subscript close parentheses V to the power of 2 end exponent plus fraction numerator 1 over denominator 2 end fraction k x to the power of 2 end exponent
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    What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10 blank c m (Take g equals 9.8 blank m divided by s to the power of 2 end exponent)

    v equals square root of 2 g h end root equals square root of 2 cross times 9.8 cross times 0.1 end root equals square root of 1.96 end root equals 1.4 blank m divided by s

    What is the velocity of the bob of a simple pendulum at its mean position, if it is able to rise to vertical height of 10 blank c m (Take g equals 9.8 blank m divided by s to the power of 2 end exponent)

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    The block of mass M moving on the frictionless horizontal surface collides with the spring of spring constant k and compresses it by length L. The maximum momentum of the block after collides is

    Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE.
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    fraction numerator 1 over denominator 2 end fraction k L to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M v to the power of 2 end exponent
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    Momentum would be maximum when KE would be maximum and this is the case when total elastic PE is converted KE.
    According to conservation of energy
    fraction numerator 1 over denominator 2 end fraction k L to the power of 2 end exponent equals fraction numerator 1 over denominator 2 end fraction M v to the power of 2 end exponent
    Or k L to the power of 2 end exponent equals fraction numerator open parentheses M v close parentheses to the power of 2 end exponent over denominator M end fraction
    M K L to the power of 2 end exponent equals p to the power of 2 end exponent left parenthesis p equals M v right parenthesis
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    The area covered by the curve of V-t graph and time axis is equal to magnitude of

    The area covered by the curve of V-t graph and time axis is equal to magnitude of

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    Let mass A moves with velocity v and collides inelastically with mass B comma which is at rest

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    Final horizontal momentum of system
    (after collision) equals m V cos invisible function application theta ….(ii)
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    m v equals m V cos invisible function application theta rightwards double arrow v equals V cos invisible function application theta …(iii)
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    fraction numerator m v over denominator square root of 3 end fraction minus m V sin invisible function application theta equals 0 rightwards double arrow fraction numerator v over denominator square root of 3 end fraction equals V sin invisible function application theta …(iv)
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    physics-General
    Let mass A moves with velocity v and collides inelastically with mass B comma which is at rest

    According to problem mass A moves in a perpendicular direction and let the mass B moves at angle theta with the horizontal with velocity v
    Initial horizontal momentum of system
    (before collision) equals m v ….(i)
    Final horizontal momentum of system
    (after collision) equals m V cos invisible function application theta ….(ii)
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    m v equals m V cos invisible function application theta rightwards double arrow v equals V cos invisible function application theta …(iii)
    Initial vertical momentum of system (before collision) is zero
    Final vertical momentum of system fraction numerator m v over denominator square root of 3 end fraction minus m V sin invisible function application theta
    From the conservation of vertical linear momentum
    fraction numerator m v over denominator square root of 3 end fraction minus m V sin invisible function application theta equals 0 rightwards double arrow fraction numerator v over denominator square root of 3 end fraction equals V sin invisible function application theta …(iv)
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