Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 3 (Q.No. 2)
2.
The loss of kinetic energy during inelastic impact, is given by(where m1 = Mass of the first body,m2 = Mass of the second body, and u1 and u2 = Velocities of the first and second bodies respectively.)
Discussion:
7 comments Page 1 of 1.
Tanmoy said:
9 years ago
m1u1 + m2u2 = (m1 + m2)v -> Find v.
initial KE = 1/2(m1(u1^2)) + 1/2(m2u2^2) Final KE = 1/2(m1 + m2)v^2, substitute v from momentum eqn.
Initial KE - final KE = the required solution.
initial KE = 1/2(m1(u1^2)) + 1/2(m2u2^2) Final KE = 1/2(m1 + m2)v^2, substitute v from momentum eqn.
Initial KE - final KE = the required solution.
(2)
Raj said:
6 years ago
For inelastic bodies, e=0.
The loss in Kinetic energy during elastic impact is.
E = m1'm2' (u1-u2) ^2' (1-e^2) /2 (m1+m2).
So here e =0 so answer is A.
The loss in Kinetic energy during elastic impact is.
E = m1'm2' (u1-u2) ^2' (1-e^2) /2 (m1+m2).
So here e =0 so answer is A.
(1)
Gopi said:
6 years ago
For inelastic bodies "e" value varies from 0 and 1.
Akhil said:
1 decade ago
Any one who can explain it please?
Mouli said:
4 years ago
Explain the answer.
Ekta said:
1 decade ago
Explain please?
Nazi said:
1 decade ago
Why? Option A.
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