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The reflecting surface of a plane mirror is vertical. A particle is projected in a vertical plane which is also perpendicular to plane of the mirror . The initial velocity of the particle is 10m/s and the angle of projection is 60 degree.The point of projection is at a distance 5m from the mirror .The particle moves towards the mirror .Just before the particle touches the mirror the velocity of approach of the particle and its image is

a)10m/s

b)5m/s

c)10\sqrt{}3 m/s

d)5\sqrt{}3 m/s

Answers (2)

Acceleration =\frac{v_{1}}{t_{1}}

\therefore    velocity at time t=(\frac{v_{1}}{t_{1}})t

Power delivered at time t

    = F. v

    = ma .v 

P=m(\frac{v_{1}}{t_{1}}).(\frac{v_{1}}{t_{1}})t

P=m\frac{v_{1}^{2}}{t_{1}^{2}} t

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Safeer PP

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any instant only the component of velocity perpendicular to the mirror matters as parrallel component will be same for both image and object . the perpendicular component is constant= ucos60=5m/s

so the object and image are moving towards each other at a speed of 5m/s

the velocity of approach=10m/s

2)A body of mass accelerates uniformly from rest   to v_{1}  in time t_{1}  The instantaneous power delivered to the body as a function of time t is

answer for this question is given below

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Safeer PP

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