Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

JEE: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

The document Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE

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

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

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Ans: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Sol: Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE (∵ α = 0°)
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE [as α = 0°]
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Ans:Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Sol:
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
or Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Initial velocity of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
∵ Initial velocity of 1st particle w.r.t. that of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
(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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ Horizontal velocity of 1st particle w.r.t. that of 2nd particle
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
∴ 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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
(b) Dotted path [(in lift) acceleration upwards]
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
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 Notes | Study DC Pandey Solutions for JEE Physics - JEE
Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE
or (u1 sin θ1 - u2 sin θ2) t = 10 …(ii)

The document Projectile Motion JEE MAin(Part - 2) - Physics, Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE is a part of the JEE Course DC Pandey Solutions for JEE Physics.
All you need of JEE at this link: JEE

Related Searches

Exam

,

Projectile Motion JEE MAin(Part - 2) - Physics

,

Summary

,

Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

,

Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

,

shortcuts and tricks

,

Extra Questions

,

Important questions

,

ppt

,

study material

,

MCQs

,

video lectures

,

practice quizzes

,

Sample Paper

,

Projectile Motion JEE MAin(Part - 2) - Physics

,

mock tests for examination

,

Free

,

Viva Questions

,

Projectile Motion JEE MAin(Part - 2) - Physics

,

Semester Notes

,

past year papers

,

Solution by DC Pandey Notes | Study DC Pandey Solutions for JEE Physics - JEE

,

Objective type Questions

,

pdf

,

Previous Year Questions with Solutions

;