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3 Apr 2009 21:56:48 IST

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COLLISION................ COMPREHENSION.............. CONCEPTUAL

Kinematics , NewtonsLawsCirMotion , Work Power Energy collisions , Rotation , Gravitation , Simple Harmonic Motion , Fluid Mechanics , Materials Wave Sound , Super Position , units dimensions , com momentum , shm , wave sound

Q38) Comprehension for 38-40

Any body thrown at some angle on the earth surface follows a parabolic path.
The motion of the body is given as
y=xtanθ₀-gx²/2(v₀cosθ₀)²
where θ₀ is the angle of projection and v₀ is the velocity of projection
coefficient of restitution is defined as the ratio of velocity of separation and velocity of approach.
Let there be two bodies 1 and 2. For 1 coefficient of restitution is 3/5 and for 2 it is 1
Both the bodies are projected at same angle and with same velocity. The bodies continue to move after rebounding.
velocity of of the bodies is given as v₀=15i+20j m/s
take g=10 m/s²

Find the time after which body 1 will stop

a) 4 sec b) 2.5 sec c) 12s d) none of these

Q39) The Distance between the bodies when the body 1 stops is

a) 0 b) 10 m c) 20m d) none of these

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Comments (3)

mita

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Joined: 25 Jan 2009

Posts: 46

4 Apr 2009 13:30:59 IST

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use conocept of infinty.........

VARUN RAJ

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Joined: 16 Mar 2008

Posts: 1825

4 Apr 2009 20:15:23 IST

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i fail to understand how there can be 2 different e

Prakhar Banga

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4 Apr 2009 20:36:24 IST

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The first body has velocities as 15i+20j m/s, 15i+(3/5)20j m/s, 15i+(3/5)^2 20j m/s....... during its motion.

The ranges for successive collisions are given as 2v(hor.) v(ver.) / g = 2v(hor.) v(ver.) / 10 .

Adding 2(15)(20) / 10 + 2(3/5)(15)(20) / 10 + 2(3/5)^2(15)(20) / 10................. infinity = 2(15)(20)/10 (1/ (1-3/5)) = 150 metres, which is the final range of particle 1.

The times of flight for successive collisions are given as 2v(ver.) / g

Adding the times of flight as 2(20) / 10 + 2(20)(3/5) / 10 .... infinity = 2(20) / 10 (1/(1-(3/5))) = 10 seconds. Option D.

There is no horizontal gap between the two bodies, because their horizontal velocities are the same.

The time of flight of the second particle is 2(20)/(10) = 4 seconds

At t = 10, the second body will be at its maximum height(It achieves max. height at 2, 6, 10..... seconds)

So the body 1 will be at the ground and body 2 will be at v(ver)^2 / 2g = (20)^2 / 2(10) = 20 metres. Option C.

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