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Work Energy and Power Test - 58

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Work Energy and Power Test - 58
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  • Question 1
    1 / -0
    Two bodies traveling with velocity 20ms120ms^{-1} and 40ms140ms^{-1} have same K.E. If the mass of the body having velocity 40ms140ms^{-1} is 8 kg the mass of other body is 
    Solution
    Given
    V1=20m/s V_1 = 20 m/s  and  V2=40m/s V_2 = 40 m/s
    M2=8 kg M_2 = 8\ kg

    K.E1=K.E2K.E_1 = K.E_2
    12M1V12=12M2V22\dfrac{1}{2}M_1V_1^2 = \dfrac{1}{2}M_2V_2^2
        M1=M2×V22V12=8×402202=32 kg\implies M_1 = M_2 \times \dfrac{V_2^2}{V_1^2} = 8\times \dfrac{40^2}{20^2} = 32\ kg
  • Question 2
    1 / -0
    A body of mass 2Kg\mathrm { Kg } moving (initially) with 10m/s\mathrm { m } / \mathrm { s } is acting upon by a resultant constar which is opposite to its initial velocity. Its speed decreases to 4m/s\mathrm { m } / \mathrm { s } in 1 s. Then the force 
  • Question 3
    1 / -0
    A body of mass 1 kg falls from rest through the air, a distance of 200 m and acquires a speed 50 m/s. Work done against air friction is 
    Solution

  • Question 4
    1 / -0
     A block of mass m is oscillating on smooth between two light springs of spring constant K separated by a distance I colliding elastically with the spring.If the velocity of the blocks is increased by an external impulse when it is not touching either of the spring then time period.

  • Question 5
    1 / -0
    A perfectly elastic ball falls on a horizontal floor from a height in a time tt. It will hit the floor again after a time tt'. The ratio of tt' and t is 
  • Question 6
    1 / -0
    As shown in figure, ball of of mass mkgm \,kg  is suspended by a string cm\ell \,cm long. Keeping the string always taut, the ball describes a horizontal circle of radius 2\dfrac{\ell}{2}. Calculate the angular speed.

    Solution
    Let θ\theta be the angle made by the string and with the vertical 
        tanθ=13\implies tan \theta= \dfrac1 {\sqrt{3}}  
    Let TT be the tension in the string and ω\omega is the angular speed.

    From Free Body Diagram 
     Tcosθ=mgT cos \theta =mg
    Tsinθ=mω2l2T sin \theta =m\omega^2 \dfrac l2

    Form above two equations we get 

    ω=2tanθgl\omega =\sqrt{\dfrac{2tan \theta g}{l}}
    using the value of tanθ tan \theta

    Angular speed , ω= 2g3l\omega= \sqrt{\dfrac{2 g}{\sqrt{3}l}}
  • Question 7
    1 / -0
    A man of mass 20 kg is running in a straight line with velocity 10 m/s. What is value of his kinetic energy?
    Solution
    The kinetic energy of the particle is given as-
    K.E.=12mv2K.E. = \dfrac{1}{2}mv^2
    mm; mass of the object
    vv; speed of the object
    K.E.=12×20×102J\Rightarrow K.E. = \dfrac{1}{2} \times 20 \times 10^2J
    K.E.=1000J\Rightarrow K.E. = 1000 J
  • Question 8
    1 / -0
    A body of mass 8 kg8\ kg collides elastically with a stationary mass of 2 kg2\ kg. If initial KEKE of moving mass be EE, the kinetic energy left with it after the collision will be:
    Solution

  • Question 9
    1 / -0
    A rubber band of length λ\lambda has a stone of mass mm tied to its one end. It is whirled with speed vsecondv second that the stone described a horizontal circular path. The tension TT in the rubber band is -
    Solution
    As there is only Tension T is theonly force. So it is the necessary centripetal force 
     T=mv2λT= \dfrac{mv^2}{\lambda}
  • Question 10
    1 / -0
    A ball of mass m1m_{1} is moving with velocity vv. It collides head on elastically with a stationary ball of mass m2m_{2}. The velocity of ball becomes v3\dfrac{v}{3} after collision, then the value of the ratio m2m1\dfrac{m_{2}}{m_{1}} is:
    Solution

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