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JIPMER 2019
Exam
JIPMER
Year
2019
Subject
Physics
Chapter
Motion in Two Dimensions

JIPMER 2019 · Physics · Motion in Two Dimensions

A projectile of mass 1 kg is projected with a speed of 10 ~m / s at...

A projectile of mass 1 kg is projected with a speed of 10 m/s10 \mathrm{~m} / \mathrm{s} at an angle of 60∘60^{\circ} from the horizontal when projectile is at its highest point, its magnitude of angular momentum (about point of projection)

Options

  1. A752Nm/s\frac{75}{2} \mathrm{Nm} / \mathrm{s}
  2. B754Nm/s\frac{75}{4} \mathrm{Nm} / \mathrm{s}
  3. C75Nm/s75 \mathrm{Nm} / \mathrm{s}
  4. D150Nm/s150 \mathrm{Nm} / \mathrm{s}

Correct answer

B. 754Nm/s\frac{75}{4} \mathrm{Nm} / \mathrm{s}

Step-by-step solution

Given, mass of projectile, m=1kg Speed of projectile, v=10m/sθ=60∘Magnitude of angular momentum at highest point, L= mass × velocity at highest point × perpendicular distance =m× horizontal component of velocity × maximum height =1×10cos⁡60×v2sin⁡260∘2g=10×12×10×102×10×34=754Nm/s\begin{aligned} & \text{Given, mass of projectile, } \mathrm{m}=1 \mathrm{ kg} \\ & \text{ Speed of projectile, } \mathrm{v}=10 \mathrm{ m} / \mathrm{s} \theta=60^{\circ} \\ & \text{Magnitude of angular momentum at highest point, } \\ &\begin{array}{r}\mathrm{L}=\text{ mass } \times \text{ velocity at highest point } \times \text{ perpendicular distance }\end{array} \\ & =\mathrm{m}\times \text{ horizontal component of velocity } \times\text{ maximum height } \\ & =1 \times 10 \cos 60 \times \frac{\mathrm{v}^2\sin ^2 60^{\circ}}{2\mathrm{ g}}=10 \times \frac{1}{2} \times \frac{10 \times 10}{2 \times 10} \times \frac{3}{4} \\ & =\frac{75}{4} \mathrm{Nm} / \mathrm{s}\end{aligned}
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