The potential energy of a 1 kg particle free move along the x-axis is given by
straight V left parenthesis straight x right parenthesis space equals space open parentheses straight x to the power of 4 over 4 minus straight x squared over 2 close parentheses space straight J

The total mechanical energy of the particle 2 J. Then, the maximum speed (in m/s) is

  • 2

  • 3 divided by square root of 2
  • square root of 2
  • square root of 2

B.

3 divided by square root of 2 kE subscript max space equals space straight E subscript straight T minus straight U subscript min
straight U subscript min space left parenthesis plus-or-minus 1 right parenthesis space equals negative 1 fourth straight J
KE subscript max space equals space 9 divided by 4 space straight J thin space
rightwards double arrow space straight U space equals space fraction numerator 3 over denominator square root of 2 end fraction space straight J
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A spherical ball of mass 20 kg is stationary at the top of a hill of height 100 m. It rolls down a smooth surface to the ground, then climbs up another hill of height 30 m and finally rolls down to a horizontal base at a height of 20 m above the ground. The velocity attained by the ball is

  • 40 m/s

  • 20 m/s

  • 10 m/s

  • 10 m/s


A.

40 m/s

mgh space equals space 1 half mv squared
straight v equals space square root of 2 gh end root
space equals square root of 2 space straight x space 10 space straight x 80 end root space equals 40 space straight m divided by straight s
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A force straight F with rightwards arrow on top space equals space left parenthesis 5 straight i with hat on top space plus 3 straight j with hat on top space plus 2 straight k with hat on top right parenthesis space straight N is applied over a particle which displaces it from its origin to the straight r with rightwards arrow on top space equals space left parenthesis 2 straight i with hat on top minus straight j with hat on top right parenthesis space straight m.The work done on the particle in joules is

  • -7

  • +7

  • +10

  • +10


B.

+7

Work done in displacing the particle
straight W space equals space straight F with rightwards arrow on top space. straight r with rightwards arrow on top
space equals space left parenthesis 5 straight i with hat on top space plus space 3 straight j with hat on top space plus 2 straight k with hat on top right parenthesis. left parenthesis 2 straight i with hat on top space minus straight j with hat on top right parenthesis
space equals space 5 space straight x space 2 space plus 3 space straight x space left parenthesis negative 1 right parenthesis space plus space 2 straight x 0
space equals space 10 minus 3
equals space 7 straight J

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A body of mass 2m moving with velocity v makes a head on elastic collision with another body of mass m which is initially at rest. Loss of kinetic energy of the colliding body (mass 2m) is

  • 1/9  of its initial kinetic energy

  • 1/6  of its initial kinetic energy 

  • 8/9 of its initial kinetic energy

  • 1/2 of its initial kinetic energy


C.

8/9 of its initial kinetic energy

Initial K.E of ball of mass 2m = K1

        = 12 × 2m × v2= mv2

Collision is elastic so both K.E and momentum are conserved. Let velocities of balls are v1 and v2 after collision

       

So, KE is conserved

 12 2mv2 = 12 2mv12 + 12 mv22  v2 = v12 + 12 v22              ......... (i)

And, momentum is conserved

⇒ (2m)v + m(0) = 2m (v1) + mv2

⇒                  2v = 2v1 + v2        ........ (ii)

Put this value in Eq. (i), we get

         v2 = v12 + 12 × 4 v - v12 3v12 - 4vv1 + v2 = 0 3 v1v2 - 4 v1v + 1 = 0or    v1v = - - (-4) ± 16 - 122 × 3 v1v  = 4 ± 22 × 3    v1 = v (Not possible)or      v1 = 13 v

So, final K.E of ball of mass 2m,

   k2 = 12 2m v12 = 12 × 2m × v29 = 19 k1

Hence, loss of K.E. of 1st ball

    = K1 - 89 K1 = 89 K1


A uniform chain of length 2 m is kept on a table such that a length of 60 cm hangs freely from the edge of the table. The total mass of the chain is 4 kg. What is the work done in pulling the entire chain on the table?

  • 7.2 J

  • 3.6 J

  • 120 J

  • 120 J


B.

3.6 J

Mass per length
= M/L
= 4/2 = 2 kg/m
The mass of 0.6 m of chain = 0.6 x 2 = 1.2 kg
The centre of mass of hanging part = 0.6 +0 /2 = 0.3 m
Hence, work done in pulling the chain on the table
W =mgh
= 1.2 x 10 x 0.3
= 3.6 J

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