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 2 of 4.
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.
Melkamu said:
6 years ago
Thanks all for explaining.
Jitendra Rajak said:
9 years ago
As K. E is same.
Therefore, 1/2m1v1^2=1/2m2v2^2.
i.e M1v1/v1=m2v2/v2.
Now as momentum is same.
Therefore, 1/v1=1/v2.
i.e V1=v2=v (let).
Therefore, m1v > m2v, as m1 > m2 (Proved).
Therefore, 1/2m1v1^2=1/2m2v2^2.
i.e M1v1/v1=m2v2/v2.
Now as momentum is same.
Therefore, 1/v1=1/v2.
i.e V1=v2=v (let).
Therefore, m1v > m2v, as m1 > m2 (Proved).
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.
Mohsin Ali said:
3 years ago
You are absolutely right @Johny.
Ajay Dubile said:
10 months ago
Good, thanks all.
Pawan kumar said:
9 years ago
K.E1= K.E2.
(1/2)m1v1^2 = (1/2)m2v2^2.
(equation showing the equal kinetic energy where m1>m2).
By cancelling (1/2) on both sides, we get,
m1v1^2 = m2v2^2.
p1v1 = p2v2 = constant (denoting p = mv which is momentum).
p1/p2= v2/v1.
You have to know that ratio of v2/v1 should be greater than 1.
So,
m1v1/m2v2=some value greater than 1.
m1v1=some value greater than 1 x m2v2.
So m1v1 will be greater than m2v2 by the value of the ratio of v2/v1.
(1/2)m1v1^2 = (1/2)m2v2^2.
(equation showing the equal kinetic energy where m1>m2).
By cancelling (1/2) on both sides, we get,
m1v1^2 = m2v2^2.
p1v1 = p2v2 = constant (denoting p = mv which is momentum).
p1/p2= v2/v1.
You have to know that ratio of v2/v1 should be greater than 1.
So,
m1v1/m2v2=some value greater than 1.
m1v1=some value greater than 1 x m2v2.
So m1v1 will be greater than m2v2 by the value of the ratio of v2/v1.
Lokesh said:
1 decade ago
K.E. = 1/2mv^2.
.
. . v1 = v2.
Momentum = mv.
So, m1v1 > m2v2.
.
. . v1 = v2.
Momentum = mv.
So, m1v1 > m2v2.
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