General
Easy
Physics-

A boy throws a cricket ball from the boundary to the wicket-keeper. If the frictional force due to air cannot be ignored, the forces acting on the ball at the position X are respected by

Physics-General

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    Answer:The correct answer is: The forces acting on the ball will be (i) in the direction opposite to its motion i e comma frictional force and(ii) weight m g.

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    Related Questions to study

    General
    physics-

    Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

    R=u2sin2θg=2uxvyg Range horizontal initial velocity (ux) In path 4 range is maximum so football possess maximum horizontal velocity in the path    

    Figure shows four paths for a kicked football. Ignoring the effects of air on the flight, rank the paths according to initial horizontal velocity component, highest first

    physics-General
    R=u2sin2θg=2uxvyg Range horizontal initial velocity (ux) In path 4 range is maximum so football possess maximum horizontal velocity in the path    
    General
    physics-

    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction

    A body of mass m is moving with a uniform speed v along a circle of radius r, what is the average acceleration in going from A to B?

    physics-General
    Here, T equals fraction numerator 2 pi r over denominator 4 v end fraction equals fraction numerator pi r over denominator 2 v end fraction
    Change in velocity is going from A to B = v square root of 2
    Average acceleration equals fraction numerator v square root of 2 over denominator pi r divided by 2 v end fraction equals fraction numerator 2 square root of 2 v to the power of 2 end exponent over denominator pi r end fraction
    General
    physics-

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis
    fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    5 equals 20 minus 8 plus 8 cos invisible function application theta
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity V at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    physics-General
    V to the power of 2 end exponent equals U to the power of 2 end exponent minus 2 g left parenthesis L minus L cos invisible function application theta right parenthesis
    fraction numerator 5 g L over denominator 4 end fraction equals 5 g L minus 2 g L left parenthesis 1 minus cos invisible function application theta right parenthesis

    5 equals 20 minus 8 plus 8 cos invisible function application theta
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    fraction numerator 3 pi over denominator 4 end fraction less than theta less than pi
    General
    physics-

    A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

    At A comma blank v subscript A end subscript equals square root of g l end root
    At B comma blank v subscript B end subscript equals square root of 5 g l end root
    and at D comma blank v subscript D end subscript equals square root of 3 g l end root
    Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
    Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
    So, D comma theta equals 90 degree
    therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g

    A particle of mass m attracted with a string of length l is just revolving on the vertical circle without slacking of the string. If v subscript A end subscript comma v subscript B end subscript and v subscript D end subscript are speed at position A comma B and D then

    physics-General
    At A comma blank v subscript A end subscript equals square root of g l end root
    At B comma blank v subscript B end subscript equals square root of 5 g l end root
    and at D comma blank v subscript D end subscript equals square root of 3 g l end root
    Thus, v subscript B end subscript greater than v subscript D end subscript greater than v subscript A end subscript
    Also, T equals 3 blank m g left parenthesis 1 plus cos invisible function application theta right parenthesis
    So, D comma theta equals 90 degree
    therefore blank T equals 3 blank m g open parentheses 1 plus theta close parentheses equals 3 blank m g
    General
    physics-

    A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, v2v1 should be

    Equating velocities along the vertical, v2=v1sin30°or   v2v1=12    

    A projectile A is thrown at an angle of 30° to the horizontal from point P. At the same time, another projectile B is thrown with velocity v2 upwards from the point Q vertically below the highest point. For B to collide with A, v2v1 should be

    physics-General
    Equating velocities along the vertical, v2=v1sin30°or   v2v1=12    
    General
    physics-

    In a two dimensional motion of a particle, the particle moves from point A, position vector stack r with rightwards arrow on top subscript 1 end subscript. If the magnitudes of these vectors are respectively, r subscript 1 end subscript=3 and r subscript 2 end subscript equals 4 and the angles they make with the x-axis are theta subscript 1 end subscript equals 75 degree and 15degree, respectively, then find the magnitude of the displacement vector

    Displacement equals A B
    angle between stack r subscript 1 end subscript with rightwards arrow on top and stack r subscript 2 end subscript with rightwards arrow on top
    theta equals 75 degree minus 15 degree equals 60 degree
    From figure
    A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta
    equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4 cos60 degree
    equals 13
    A B equals square root of 13

    In a two dimensional motion of a particle, the particle moves from point A, position vector stack r with rightwards arrow on top subscript 1 end subscript. If the magnitudes of these vectors are respectively, r subscript 1 end subscript=3 and r subscript 2 end subscript equals 4 and the angles they make with the x-axis are theta subscript 1 end subscript equals 75 degree and 15degree, respectively, then find the magnitude of the displacement vector

    physics-General
    Displacement equals A B
    angle between stack r subscript 1 end subscript with rightwards arrow on top and stack r subscript 2 end subscript with rightwards arrow on top
    theta equals 75 degree minus 15 degree equals 60 degree
    From figure
    A B to the power of 2 end exponent equals r subscript 1 end subscript superscript 2 end superscript plus r subscript 2 end subscript superscript 2 end superscript minus 2 r subscript 1 end subscript r subscript 2 end subscript c o s theta
    equals 3 to the power of 2 end exponent plus 4 to the power of 2 end exponent minus 2 cross times 3 cross times 4 cos60 degree
    equals 13
    A B equals square root of 13
    General
    physics-

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    Velocity of the bob at the point A
    v equals square root of 5 g L end root(i)
    open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals v to the power of 2 end exponent minus 2 g h open parentheses i i close parentheses
    h equals L left parenthesis 1 minus cos invisible function application theta right parenthesis left parenthesis i i i right parenthesis
    S o l v i n g blank E q s. open parentheses i close parentheses comma blank open parentheses i i close parentheses a n d blank open parentheses i i i close parentheses comma blank w e blank g e t
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    o r blank theta equals c o s to the power of negative 1 end exponent open parentheses negative fraction numerator 7 over denominator 8 end fraction close parentheses equals 151 degree

    A bob of mass M is suspended by a massless string of length L. The horizontal velocity v at position A is just sufficient to make it reach the point B. The angle theta at which the speed of the bob is half of that at A, satisfies

    physics-General
    Velocity of the bob at the point A
    v equals square root of 5 g L end root(i)
    open parentheses fraction numerator v over denominator 2 end fraction close parentheses to the power of 2 end exponent equals v to the power of 2 end exponent minus 2 g h open parentheses i i close parentheses
    h equals L left parenthesis 1 minus cos invisible function application theta right parenthesis left parenthesis i i i right parenthesis
    S o l v i n g blank E q s. open parentheses i close parentheses comma blank open parentheses i i close parentheses a n d blank open parentheses i i i close parentheses comma blank w e blank g e t
    cos invisible function application theta equals negative fraction numerator 7 over denominator 8 end fraction
    o r blank theta equals c o s to the power of negative 1 end exponent open parentheses negative fraction numerator 7 over denominator 8 end fraction close parentheses equals 151 degree
    General
    physics-

    A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P blanktoblank Q blankisblank left parenthesis angle P O Q equals 40 degree right parenthesis

    Change in velocity equals 2 v sin invisible function application open parentheses theta divided by 2 close parentheses equals 2 v sin invisible function application 20 degree

    A particle is moving on a circular path of radius r with uniform velocity v. The change in velocity when the particle moves from P blanktoblank Q blankisblank left parenthesis angle P O Q equals 40 degree right parenthesis

    physics-General
    Change in velocity equals 2 v sin invisible function application open parentheses theta divided by 2 close parentheses equals 2 v sin invisible function application 20 degree
    General
    physics-

    A fighter plane enters inside the enemy territory, at time t equals 0 with velocity v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent and moves horizontally with constant acceleration a equals 20 m s to the power of negative 2 end exponent (see figure). An enemy tank at the border, spot the plane and fire shots at an angle theta equals 60 degree with the horizontal and with velocity u equals 600 blank m s to the power of negative 1 end exponent. At what altitude H of the plane it can be hit by the shot?

    If it is being hit then
    d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t
    or t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

    therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s
    H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent
    equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25
    H equals 2473 blank m

    A fighter plane enters inside the enemy territory, at time t equals 0 with velocity v subscript 0 end subscript equals 250 blank m s to the power of negative 1 end exponent and moves horizontally with constant acceleration a equals 20 m s to the power of negative 2 end exponent (see figure). An enemy tank at the border, spot the plane and fire shots at an angle theta equals 60 degree with the horizontal and with velocity u equals 600 blank m s to the power of negative 1 end exponent. At what altitude H of the plane it can be hit by the shot?

    physics-General
    If it is being hit then
    d equals v subscript 0 end subscript t plus fraction numerator 1 over denominator 2 end fraction a t to the power of 2 end exponent equals left parenthesis u cos invisible function application theta right parenthesis t
    or t equals fraction numerator u cos invisible function application theta minus v subscript 0 end subscript over denominator a divided by 2 end fraction

    therefore blank t equals fraction numerator 600 cross times fraction numerator 1 over denominator 2 end fraction minus 250 over denominator 10 end fraction equals 5 blank s
    H equals open parentheses u sin invisible function application theta close parentheses t minus fraction numerator 1 over denominator 2 end fraction cross times g t to the power of 2 end exponent
    equals 600 cross times fraction numerator square root of 3 over denominator 2 end fraction cross times 5 minus fraction numerator 1 over denominator 2 end fraction cross times 10 cross times 25
    H equals 2473 blank m
    General
    physics-

    From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure, which of the following statements is/are correct?

    H e r e comma blank alpha equals 2 theta comma blank beta equals theta

    T i m e blank o f blank f l i g h t blank o f blank A blank i s comma
    T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction
    equals fraction numerator 2 u sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator g cos invisible function application theta end fraction
    equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    T i m e blank o f blank f l i g h t blank o f blank B blank i s comma blank T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction
    equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towardsA B, therefore collision will take place between the two in mid air.

    From an inclined plane two particles are projected with same speed at same angle theta, one up and other down the plane as shown in figure, which of the following statements is/are correct?

    physics-General
    H e r e comma blank alpha equals 2 theta comma blank beta equals theta

    T i m e blank o f blank f l i g h t blank o f blank A blank i s comma
    T subscript 1 end subscript equals fraction numerator 2 u sin invisible function application open parentheses alpha minus beta close parentheses over denominator g cos invisible function application beta end fraction
    equals fraction numerator 2 u sin invisible function application open parentheses 2 theta minus theta close parentheses over denominator g cos invisible function application theta end fraction
    equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    T i m e blank o f blank f l i g h t blank o f blank B blank i s comma blank T subscript 2 end subscript equals fraction numerator 2 u sin invisible function application theta over denominator g cos invisible function application theta end fraction
    equals fraction numerator 2 u over denominator g end fraction tan invisible function application theta
    So, T subscript 1 end subscript equals T subscript 2 end subscript. The acceleration of both the particles is g downwards. Therefore, relative acceleration between the two I zero or relative motion between the two is uniform. The relative velocity of A w.r.t. B is towardsA B, therefore collision will take place between the two in mid air.
    General
    physics-

    A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2 divided by pi revolutions per s e c o n d around the vertical axis through the fixed end as shown in figure , then tension in the string is

    T sin invisible function application theta equals M omega to the power of 2 end exponent R(i)
    T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta(ii)
    From (i) and (ii)
    T equals M omega to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L

    A string of length L is fixed at one end and carries a mass M at the other end. The string makes 2 divided by pi revolutions per s e c o n d around the vertical axis through the fixed end as shown in figure , then tension in the string is

    physics-General
    T sin invisible function application theta equals M omega to the power of 2 end exponent R(i)
    T sin invisible function application theta equals M omega to the power of 2 end exponent L sin invisible function application theta(ii)
    From (i) and (ii)
    T equals M omega to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent n to the power of 2 end exponent L
    equals M blank 4 pi to the power of 2 end exponent open parentheses fraction numerator 2 over denominator pi end fraction close parentheses to the power of 2 end exponent L equals 16 blank M L
    General
    physics-

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction

    A piece of wire is bent in the shape of a parabola y equals k x to the power of 2 end exponent blank left parenthesis y-axis vertical) with a bead of mass m on it. The bead can side on the wire without friction. It stays at the lowest point of the parabola when the wire is at rest. The wire is now accelerated parallel to the x-axis with a constant acceleration a. The distance of the new equilibrium position of the bead, where the bead can stay at rest with respect to the wire, from the y-axis is

    physics-General
    m a cos invisible function application theta equals m g cos invisible function application left parenthesis 90 minus theta right parenthesis
    rightwards double arrow fraction numerator a over denominator g end fraction equals tan invisible function application theta rightwards double arrow fraction numerator a over denominator g end fraction equals fraction numerator d y over denominator d x end fraction
    rightwards double arrow fraction numerator d over denominator d x end fraction open parentheses k x close parentheses to the power of 2 end exponent equals fraction numerator a over denominator g end fraction rightwards double arrow x equals fraction numerator a over denominator 2 g k end fraction
    General
    physics-

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m

    A frictionless track A B C D E ends in a circular loop of radius R comma figure. A body slides down the track from point A which is at a height h equals 5 blankcm. Maximum value of R for the body to successfully complete the loop is

    physics-General
    For successfully completing the loop,
    h equals fraction numerator 5 over denominator 4 end fraction R rightwards double arrow R equals fraction numerator 2 h over denominator 5 end fraction equals fraction numerator 2 cross times 5 over denominator 5 end fraction equals 2 c m
    General
    physics-

    The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through R, where P R equals

    Equating the moments about R
    6 cross times P R equals 4 cross times R Q
    P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q

    The resultant of a system of forces shown in figure is a force of 10 N parallel to given forces through R, where P R equals

    physics-General
    Equating the moments about R
    6 cross times P R equals 4 cross times R Q
    P R equals fraction numerator 4 over denominator 6 end fraction R Q equals fraction numerator 2 over denominator 3 end fraction R Q
    General
    physics-

    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

    Horizontal component of velocity at A

    v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction
    A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction

    The time taken by the projectile to reach from A to B is t comma then the distance A B is equal to

    physics-General
    Horizontal component of velocity at A

    v subscript H end subscript equals u cos invisible function application 60 degree equals fraction numerator u over denominator 2 end fraction blank therefore A C equals u subscript H end subscript cross times t equals fraction numerator u t over denominator 2 end fraction
    A B equals A C sec invisible function application 30 degree equals fraction numerator u t over denominator 2 end fraction cross times fraction numerator 2 over denominator square root of 3 end fraction equals fraction numerator u t over denominator 2 end fraction