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A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2?
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A particle of mass 1 kg is projected with an initial velocity of 2√5 m...
Problem Statement
A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal. At the same instant, another particle of the same mass is projected vertically upwards from a distance 1 metre from the point of projection of the first particle. If the first particle follows a straight-line path after collision with particle 2, then find the range of particle 1 and 2, coefficient of restitution for collision, and velocity of particle 2.
Solution
Finding the Range of Particle 1
Initial horizontal velocity of particle 1 can be found as:
ux = u cos θ
Where u is the initial velocity and θ is the angle of projection. Substituting the given values:
ux = 2√5 cos 45 = 2√5 / 2 = √5 m/s
The time taken by the particle to reach its maximum height can be found using:
t = u sin θ / g
Where g is the acceleration due to gravity. Substituting the given values:
t = 2√5 sin 45 / 9.8 = √5 / 4.9 s
The maximum height reached by the particle can be found using:
h = uy² / 2g
Where uy is the vertical component of initial velocity. Substituting the given values:
h = (2√5 sin 45)² / (2 × 9.8) = 5 / 2 m
The time taken by the particle to reach the ground can be found using:
t = 2tmax height = 2 × √5 / 4.9 s
The range of particle 1 can be found using:
R = ux × t = √5 × 2 × √5 / 4.9 = 2.04 m
Finding the Range of Particle 2
The time taken by particle 2 to reach the ground can be found using:
t = √(2h / g) = √(2 / 4.9) = √(20 / 49) s
The range of particle 2 can be found using:
R = 0
Since particle 2 is projected vertically upwards and it falls back at the same point of projection.
Finding the Coefficient of Restitution
The coefficient of restitution can be found using:
e = (v2f - v1f) / (u1 - u2)
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A particle of mass 1 kg is projected with an initial velocity of 2√5 m...
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A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2?
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A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2?.
A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A particle of mass 1 kg is projected with an initial velocity of 2√5 m/s at an angle of 45 degree with the horizontal .at same instant another particle ofsame mass is projected vertically upwards from at a distance 1 metre from the point of projection of first particle. if particle when follows as straight line path after collision with particle 2 then Find Range of particle 1 and 2 Coefficient of restitution for collision Velocity of particle 2?.
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