Civil Engineering - Applied Mechanics - Discussion
Discussion Forum : Applied Mechanics - Section 5 (Q.No. 4)
4.
A ball of mass 1 kg moving with a velocity of 2 m/sec collides a stationary ball of mass 2 kg and comes to rest after impact. The velocity of the second ball after impact will be
Discussion:
3 comments Page 1 of 1.
Akki said:
4 years ago
Absolutely right, Thanks @Tanvir.
Tanvir said:
5 years ago
W1.U1 + W2.U2 = W1.V1 + W2.V2
1x 2 + 2 x 0 = 1x 0 + 2 x V2.
2V2 = 2.
V2 =1 m/s.
1x 2 + 2 x 0 = 1x 0 + 2 x V2.
2V2 = 2.
V2 =1 m/s.
Asay said:
8 years ago
m1v1 = m2v2.
v2 = m1v1/m2,
v2 = 1*2/2 = 1m/sec,
v2 = 1.0m/sec.
v2 = m1v1/m2,
v2 = 1*2/2 = 1m/sec,
v2 = 1.0m/sec.
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