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.
equal to
less than
greater than
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
33 comments Page 1 of 4.

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.
(2)

Gaurav said:   6 months ago
Thanks all.
(1)

Hassan rizwi said:   8 years ago
Simple greater the mass greater the momentum.
(1)

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.
(1)

Ajay Dubile said:   2 years ago
Good, thanks all.
(1)

Harish said:   1 decade ago
@Suresh.

Given, K.E1 = K.E2, also m1>m2, K.E1 = 1/2m1v1^2, K.E2 = 1/2m2v2^2.

In order to get both the KE equal, their Velocity should be equal since m1>m2 (given). Therefore, Momentum, m1v1 > m2v2.
(1)

Deep said:   1 decade ago
1/2 m1 v1^2 = 1/2 m2 v2^2.

As m1>m2 So v2 must be >v1.

m1.v1/m2.v2=v2/v1 will have value greater than 1. So m1.v1>m2.v2.
(1)

Suresh kumawat said:   1 decade ago
How can you say that v1 = v2.
(1)

Mohsin Ali said:   4 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.


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