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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Physics
Chapter
Center of Mass, Momentum and Collision

NTA Abhyas 2020 · Physics · Center of Mass, Momentum and Collision

A particle of mass m is projected with a velocity v _ 0 at an angle...

A particle of mass mm is projected with a velocity v0{v}_{0} at an angle α\alpha with the horizontal. The coefficient of restitution for any of its impact with the smooth ground is ee. The time after which the particle will stop bouncing off the ground is

A particle of mass m is projected with a velocity v _ 0 at an angle α with the horizontal. The coefficient of

Options

  1. AT=2v0sinα(1+e)g(1−e)T = \dfrac{2{v}_{0}sin\alpha ( 1 + e )}{g ( 1 - e )}
  2. BT=2v0sinα(1−e)g(1+e)T = \dfrac{2{v}_{0}sin\alpha ( 1 - e )}{g ( 1 + e )}
  3. CT=2v0sinαg(1+e)T = \dfrac{2{v}_{0}sin\alpha}{g ( 1 + e )}
  4. DT=2v0sinαg(1−e)T = \dfrac{2{v}_{0}sin\alpha}{g ( 1 - e )}

Correct answer

D. T=2v0sinαg(1−e)T = \dfrac{2{v}_{0}sin\alpha}{g ( 1 - e )}

Step-by-step solution

After each collision, the particle loses some part of its vertical velocity while its horizontal velocity remains unchanged.

T=2v0sinαg+2(ev0)sinαg+2(e2v0)sinαg+..........T = \dfrac{2{v}_{0}sin\alpha}{g} + \dfrac{2\left(e{v}_{0}\right)sin\alpha}{g} + \dfrac{2\left(e^{2}{v}_{0}\right)sin\alpha}{g} +..........

⇒T=2v0sinαg(1+e+e2+.........)⇒ T = \dfrac{2{v}_{0}sin\alpha}{g} \left(1 + e + e^{2} +.........\right)

T=2v0sinαg(1−e)T = \dfrac{2{v}_{0}sin\alpha}{g ( 1 - e )}

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