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:
32 comments Page 3 of 4.
Jean luc I said:
8 years ago
The kinetic energy is a function of mass M and velocity V^2, so if we assume the kinetic energy to be the same for both M1 and M2 and fix the velocity such that V1=V2 the M1 = M2, that is True.
So, if we FIX the velocity such that V1=V2, and M1>M2 it is also true that the KE of Object 1> KE of object 2.
So, finally, If the KE of object 1 = KE of object 2, and M1>M2, then the V2>V1 so that object 2 can make up the remainder of the energy that object 1 gets from its mass being greater.
Since V2>V1 the V2/V1 >1.
Thus with KE = 1/2 MV^2 or PV/2.
the KE1=KE2 or P1V1/2=P2V2/2 get rid of the 1/2 since its constants then,
M1V1V1= M2V2V2 divide both sides by V1 so, M1V1 = M2V2*(V2/V1) where V2/V1 has to be greater than 1 as shown above.
So conclusion, P1=P2(V2/V1), where V2/V1 is greater than 1 nd we get P1>P2.
So, if we FIX the velocity such that V1=V2, and M1>M2 it is also true that the KE of Object 1> KE of object 2.
So, finally, If the KE of object 1 = KE of object 2, and M1>M2, then the V2>V1 so that object 2 can make up the remainder of the energy that object 1 gets from its mass being greater.
Since V2>V1 the V2/V1 >1.
Thus with KE = 1/2 MV^2 or PV/2.
the KE1=KE2 or P1V1/2=P2V2/2 get rid of the 1/2 since its constants then,
M1V1V1= M2V2V2 divide both sides by V1 so, M1V1 = M2V2*(V2/V1) where V2/V1 has to be greater than 1 as shown above.
So conclusion, P1=P2(V2/V1), where V2/V1 is greater than 1 nd we get P1>P2.
Athul said:
7 years ago
Let me explain it theoretically,
It is given that the masses m1>m2 and it attains same K.E. which means that they are moving at same speed. Inorder to attain same speed more force should be applied to mass m1 as it is greater than m2. Thus by using momentum formula we can prove that m1v1>m2v2 (momentum,P= m*v).
It is given that the masses m1>m2 and it attains same K.E. which means that they are moving at same speed. Inorder to attain same speed more force should be applied to mass m1 as it is greater than m2. Thus by using momentum formula we can prove that m1v1>m2v2 (momentum,P= m*v).
Sairohit sandela said:
7 years ago
We know KE=P^2/2m.
For equal KE(kinetic energies).
The Square of P(momentum) is proportional to the mass.
Hence greater the mass greater is the momentum.
For equal KE(kinetic energies).
The Square of P(momentum) is proportional to the mass.
Hence greater the mass greater is the momentum.
Hassan rizwi said:
7 years ago
Simple greater the mass greater the momentum.
(1)
Melkamu said:
6 years ago
Thanks all for explaining.
Pratyay said:
6 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)
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.
Mechoy said:
5 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)
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.
Kamran Ashraf said:
4 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.
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