Mechanical Engineering - Engineering Mechanics - Discussion

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.

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.

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).

(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.

(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.

1/2m1(v1)^2 = 1/2m2(v2)^2.

= m1(v1)^2 = m2(v2)^2.

= m1(v1)^2/m2(v2)^2 = 1.

= (v1)^2/(v2)^2 is greater than 1(science m1/m2 is greater than 1).

= v1^2is greater than v2^2.

= v1 is greater than v2.

So, m1v1 is greater than m2v2.

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.

@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.

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.

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).