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JEE Advanced 2012Paper 1
Exam
JEE Advanced
Year
2012
Paper
Paper 1
Subject
Physics
Chapter
Motion in Two Dimensions

JEE Advanced 2012 · Paper 1 · Physics · Motion in Two Dimensions

A small block is connected to one end of a massless spring of...

A small block is connected to one end of a massless spring of un-stretched length 4.9 m4.9 \mathrm{~m}. The other end of the spring (see the figure) is fixed. The system lies on a horizontal frictionless surface. The block is stretched by 0.2 m0.2 \mathrm{~m} and released from rest at t=0t=0. It then executes simple harmonic motion with angular frequency ω=π/3rad/s\omega=\pi / 3 \mathrm{rad} / \mathrm{s}. Simultaneously at t=0t=0, a small pebble is projected with speed vv form point PP at an angle of 45∘45^{\circ} as shown in the figure. Point PP is at a horizontal distance of 10 m10 \mathrm{~m} from OO. If the pebble hits the block at t=1 st=1 \mathrm{~s}, the value of vv is (take gg =10 m/s2=10 \mathrm{~m} / \mathrm{s}^{2} ) A small block is connected to one end of a massless spring of un-stretched length 4.9 ~m . The other end of the spring

Options

  1. A50 m/s\sqrt{50} \mathrm{~m} / \mathrm{s}
  2. B51 m/s\sqrt{51} \mathrm{~m} / \mathrm{s}
  3. C52 m/s\sqrt{52} \mathrm{~m} / \mathrm{s}
  4. D53 m/s\sqrt{53} \mathrm{~m} / \mathrm{s}

Correct answer

A. 50 m/s\sqrt{50} \mathrm{~m} / \mathrm{s}

Step-by-step solution

Time of flight of projectile, T=2Vsin⁡θgT=\frac{2 V \sin \theta}{g} ∴1=2Vsin⁡45∘g∴V=50 ms−1\therefore \quad 1=\frac{2 V \sin 45^{\circ}}{g} \quad \therefore \quad V=\sqrt{50} \mathrm{~ms}^{-1} Hence pebble is projected with a speed V=50 ms−1V=\sqrt{50} \mathrm{~ms}^{-1}
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