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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Motion in Two Dimensions

NTA Abhyas 2020 · Physics · Motion in Two Dimensions

A car is travelling at a velocity 10kmh^ - 1 on a straight road....

A car is travelling at a velocity 10kmh−110kmh^{- 1} on a straight road. The driver of the car throws a parcel with a velocity 102kmh−110\sqrt{2}kmh^{- 1}& with respect to the car, when the car is passing by a man standing on the side of the road. If the parcel is to reach the man, the direction of throw makes the following angle with the direction of motion of the carA car is travelling at a velocity 10kmh^ - 1 on a straight road. The driver of the car throws a parcel with a velocity

Options

  1. A135°135 °
  2. B45°45 °
  3. Ctan⁡−12\tan^{-1} \sqrt{2}
  4. Dtan⁡−1(12)\tan^{- 1} \left(\dfrac{1}{2}\right)

Correct answer

A. 135°135 °

Step-by-step solution

In the vector diagram, OA¯\overset{¯}{OA} and OB¯\overset{¯}{OB} represent the velocity vectors for the car and the parcel, respectively. V→R\overset{→}{V}R is the resultant velocity vector. Let θ\theta be the angle between V→car\overset{→}{V}car and V→Parcel\overset{→}{V}Parcel. Then,

A car is travelling at a velocity 10kmh^ - 1 on a straight road. The driver of the car throws a parcel with a velocity

cos⁡(180°−θ)=∣V→car∣∣V→parcel∣=10102=12\cos ( 180 ° - \theta ) = \left|\overset{→}{V}car\right|\left|\overset{→}{V}parcel\right| = \dfrac{10}{10\sqrt{2}} = \dfrac{1}{2}

180°−θ=45°180 ° - \theta = 45 °

∴θ=135°∴ \theta = 135 °
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