Q 1 :    

The bob of a pendulum was released from a horizontal position. The length of the pendulum is 10 m. If it dissipates 10% of its initial energy against air resistance, the speed with which the bob arrives at the lowest point is ______.

[Use, g = 10 ms-2]                          [2024]

  • 65 ms-1

     

  • 55 ms-1

     

  • 25 ms-1

     

  • 56 ms-1

     

(1)

=10 m,

Initial energy =mg

So, 910mg=12mv2

  910×10×10=12v2

v2=180

v=180=65 m/s



Q 2 :    

A particle is placed at the point A of a frictionless track ABC as shown in the figure. It is gently pushed towards the right. The speed of the particle when it reaches the point B is (Take g = 10 m/s2).          [2024]

  • 20 m/s

     

  • 10 m/s

     

  • 210 m/s

     

  • 10 m/s

     

(2)

By energy conservation

|ΔPE|=|ΔKE|

mg(h2-h1)=(12mv2-0)

mg[0.5]=12mv2

v=10 m/s



Q 3 :    

A disc of radius R and mass M is rolling horizontally without slipping with speed n. It then moves up an inclined smooth surface as shown in the figure. The maximum height that the disc can go up the incline is ______    [2024]

  • v2g

     

  • 34v2g

     

  • 12v2g

     

  • 23v2g

     

(3)

Rolling v0=ωR

Conservation of energy, Ki+Ui=Kf+Uf

12mv02+12[MR22]·(v0R)2+0=mgh+12[MR22]·(v02R2)

12mv02=mghmaxhmax=v022g



Q 4 :    

A bead of mass 'm' slides without friction on the wall of a vertical circular hoop of radius 'R' as shown in figure. The bead moves under the combined action of gravity and a massless spring (k) attached to the bottom of the hoop. The equilibrium length of the spring is 'R'. If the bead is released from top of the hoop with (negligible) zero initial speed, velocity of bead, when the length of spring becomes 'R', would be (spring constant is 'k', g is acceleration due to gravity)          [2025]

  • 2gR+kR2m

     

  • 2Rg+4kR2m

     

  • 2Rg+kR2m

     

  • 3Rg+kR2m

     

(4)

Work done by gravity = mg (2RR cos 60°) = 3mgR2

Work done by spring =12k(02R2)=12kR2

Net work = Change in kinetic energy

i.e.3mgR2+kR22=12mv2

or v2=3gR+kR2m

or v=3gR+kR2m



Q 5 :    

A block of mass 2 kg is attached to one end of a massless spring whose other end is fixed at a wall. The spring-mass system moves on a frictionless horizontal table. The spring's natural length is 2 m and spring constant is 200 N/m. The block is pushed such that the length of the spring becomes 1 m and then released. At distance xm (x < 2) from the wall the speed of the block will be :          [2025]

  • 10[10(2×x)3/2]m/s

     

  • 10[1(2x)2]1/2m/s

     

  • [1(2x)2]m/s

     

  • 10[1(2x)2]m/s

     

(2)

Given, Natural length of spring = 2 m

Initial compression is spring (xi)=1m

Final compression in spring (xf)=(2x)m

Using energy conservation: Ki+Ui=Kf+Uf

0+12Kxi2=12mv2+12Kxf2

12mv2=12K(xi2xf2)

12×2×v2=12×200×(12(2x)2)

v2=100[1(2x)2]

 v=10[1(2x)2]1/2