Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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Projectile Motion in Inclined Plane

Ques 15: Find time of flight and range of the projectile along the inclined plane as shown in figure, (g = 10 m/s2)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ans: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: 
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 16: Find time of flight and range of the projectile along the inclined plane as shown in figure, (g = 10m/s2)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: 
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 17: Find time of flight and range of the projectile along the inclined plane as shown in figure, (g = 10 m/s2)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev (∵ α = 0°)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev [as α = 0°]
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Ques 18: A projectile is fired with a velocity u at right angles to the slope, which is inclined at an angle θ with the horizontal. Derive an expression for the distance R to the point of impact.
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans:Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Sol:
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev

Relative Motion in Projectiles
Note:
The problems can also be solved without using the concept of relative motion.
Ques 19: A particle is projected upwards with velocity 20 m/s. Simultaneously another particle is projected with velocity 20√2 m/s at 45°. (g = 10 m/s2)
(a) What is acceleration of first particle relative to the second? 
(b) What is initial velocity of first particle relative to the other? 
(c) What is distance between two particles after 2 s?
Ans: (a) zero
(b) 20 ms-1 in horizontal direction
(c) 40 m
Sol: (a)  Acceleration of particle 1 w.r.t. that of particle 2
= (- g) - (- g)
= 0
(b) Initial velocity of 1st particle Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Initial velocity of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∵ Initial velocity of 1st particle w.r.t. that of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
(c) Horizontal velocity of 1st particle = 0 m/s
Horizontal velocity of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Horizontal velocity of 1st particle w.r.t. that of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Relative displacement of 1st particle w.r.t. 2nd particle at t = 2 s
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
∴ Distance between the particles at t = 2 s = 40 m

Ques 20: Passenger of a train just drops a stone from it. The train was moving with constant velocity. What is path of the stone as observed by 
(a) the passenger itself, 
(b) a man standing on ground?
Ans: (a) A vertical straight line
(b) A parabola
Ans: (a) As observed by passenger Vertical acceleration of stone
= g - 0 = g
Horizontal velocity of stone
= v - v = 0
∴ Path of the stone will be a straight line (downwards).
(b) As observed by man standing on ground
Vertical acceleration of stone = g
Horizontal velocity of stone = v
∴ Path of the stone will be parabolic.

Ques 21: An elevator is going up with an upward acceleration of 1 m/s2. At the instant when its velocity is 2 m/s, a stone is projected upward from its floor with a speed of 2 m/s relative to the elevator, at an elevation of 30°.
(a) Calculate the time taken by the stone to return to the floor.
(b) Sketch the path of the projectile as observed by an observer outdside the elevator.
(c) If the elevator was moving with a downward acceleration equal to g, how would the motion be altered?

Ans: (a) 0.18 s
Sol: (a) geff = g - (- a)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
(b) Dotted path [(in lift) acceleration upwards]
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Full line path [In lift at rest or moving with constant velocity upwards or downwards].
(c) If lift is moving downward with acceleration g.
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
geff = g - g = 0

Ques 21: Two particles A and B are projected simultaneously in a vertical plane as shown in figure. They collide at time t in air. Write down two necessary equations for collision to take place.
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Ans: (u1cos θ1 + u2 cos θ2) t = 20 ...(i)
(usinθ- u2 sinθ2) t = 10 ...(ii)
Sol: Horizontal motion:
x1 = u1 cos θ1 and x2 = u2 cos θ
∴ u1 cos θ1 + u2 cos θ2 = 20 …(i)
Vertical motion:
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey NEET Notes | EduRev
or (u1 sin θ1 - u2 sin θ2) t = 10 …(ii)

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