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.
Sandy said:
10 years ago
Since Momentum = Mass Acceleration.
That means Momentum is directly proportional to the mass.
So, if mass is more the momentum will also be more.
That means Momentum is directly proportional to the mass.
So, if mass is more the momentum will also be more.
Nitin said:
10 years ago
@Sashi.
But don't forget momentum also depends upon velocity and you missed that.
But don't forget momentum also depends upon velocity and you missed that.
Seshu said:
10 years ago
We know that KE = 1/2mv^2. If m value increases moment increases.
So body 1 mass is higher than body 2. So momentum of body 1 greater than body 2.
So body 1 mass is higher than body 2. So momentum of body 1 greater than body 2.
Johny said:
10 years ago
KE1 = KE2.
= 1/2M1V1^2 = 1/2M2V2^2.
= M1>M2.
= V2^2>V1^2 but V2/V1 lesser than V2^2/V1^2.
Or we say V2/V1 lesser than M1/M2 also.
So M1V1>M2V2.
= 1/2M1V1^2 = 1/2M2V2^2.
= M1>M2.
= V2^2>V1^2 but V2/V1 lesser than V2^2/V1^2.
Or we say V2/V1 lesser than M1/M2 also.
So M1V1>M2V2.
Dhiru said:
10 years ago
K.E-p^2/2m.
p - Momentum.
p - Momentum.
Amit papa said:
9 years ago
(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).
So, to make the product of two quantities equal, p1>p2.
Because v1
Because of m1>m2.
By cancelling (1/2) on both sides, we get,
m1v1^2 = m2v2^2.
p1v1 = p2v2 = constant (denoting p = mv which is momentum).
So, to make the product of two quantities equal, p1>p2.
Because v1
Because of m1>m2.
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).
Pawan kumar said:
9 years ago
(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).
So, to make the product of two quantities equal, p1>p2.
By cancelling (1/2) on both sides, we get,
m1v1^2 = m2v2^2.
p1v1 = p2v2 = constant (denoting p = mv which is momentum).
So, to make the product of two quantities equal, p1>p2.
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.
Gaurav1995 said:
9 years ago
To avoid complications assume initial values of m1 & m2 as 4 & 1 then we get velocity ratio(v2/v1) as 2 (K.E.=const) consider v1 as & v2 as 2. then m1v1=4 & m2v2 = 2.
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