A particle strikes a horizontal smooth floor with a velocity u making an angle θ with the floor and rebounds with velocity v making an angle φ with the floor. If the coefficient of restitution between the particle and the floor is e, then :
A
the impulse delivered by the floor to the body is mu(1 + e) sin θ .
B
tan φ = e tan θ .
C
v = u 
D
the ratio of the final kinetic energy to the initial kinetic energy is (cos 2 θ + e 2 sin 2 θ )
Text Solution
Verified by ExpertsThe correct answer is:
CHECK THE SOLUTION.
(a, b, c, d)
vcos φ = ucos θ
vsin φ = eusin θ
v 2 = 
v = 
∴ v = u 
tan φ = etan θ .
I = m (v LOI –u LOI )
= m (eusin θ + usin θ )
= mu (1 + e) sin θ .
k i = 
mu 2 k f = 
mv 2
= 
= cos 2 θ + e 2 sin 2 θ
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