Mechanical Engineering - Engineering Mechanics - Discussion
Discussion Forum : Engineering Mechanics - Section 1 (Q.No. 50)
50.
If two bodies having masses m1 and m2 (m1>m2) have equal kinetic energies, the momentum of body having mass m1 is __________ the momentum of body having mass m2.
Discussion:
33 comments Page 1 of 4.
Gaurav said:
3 weeks ago
Thanks all.
Ajay Dubile said:
2 years ago
Good, thanks all.
Mohsin Ali said:
3 years ago
You are absolutely right @Johny.
Kamran Ashraf said:
5 years ago
Ke1 = ke2,
m1v1^2 = m2v2^2,
m1v1 * v1 = m2v2 * v2,
Moment1 *v1 = Moment2 *v2,
Moment1/Moment2 = v2/v1.
m1>m2 so for equal KE v2>v1 i.e v2/v1 >1;
Moment1>Moment2.
m1v1^2 = m2v2^2,
m1v1 * v1 = m2v2 * v2,
Moment1 *v1 = Moment2 *v2,
Moment1/Moment2 = v2/v1.
m1>m2 so for equal KE v2>v1 i.e v2/v1 >1;
Moment1>Moment2.
Vishnu said:
5 years ago
V1 can't be equal to v2 as if they were equal how come kinetic energy be equal as mass are different? Explain please.
Mechoy said:
6 years ago
Simply.
Momentum is related to mass and velocity.
If mass and velocity increase then momentum will increase.
So finally, m1>m2.
Momentum is related to mass and velocity.
If mass and velocity increase then momentum will increase.
So finally, m1>m2.
(1)
Satya Jeet Verma said:
6 years ago
K.E= p^2/2*m.
KE1=KE2.
P1^2/2*m1=P2^/2*m2.
since m1>m2.
So P1>P2.
KE1=KE2.
P1^2/2*m1=P2^/2*m2.
since m1>m2.
So P1>P2.
Pratyay said:
7 years ago
IF m<M & v,V are their velocity respectively.
1/2(mv^2)=1/2(MV^2).
m.v*v=M.V*V.
if m<M,in order to get same K.E, v>V.
Now,
p.v=P.V (p=m.v & P=M.V).
as v>V, in order to get same K.E,
So, P>p.
1/2(mv^2)=1/2(MV^2).
m.v*v=M.V*V.
if m<M,in order to get same K.E, v>V.
Now,
p.v=P.V (p=m.v & P=M.V).
as v>V, in order to get same K.E,
So, P>p.
(2)
Melkamu said:
7 years ago
Thanks all for explaining.
Hassan rizwi said:
8 years ago
Simple greater the mass greater the momentum.
(1)
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