Question

Two bodies of masses m1 and m2 have same kinetic energy. The ratio of their momentum

Answer 1

Let the two masses m1 and m2 be denoted M and m respectively

Let Kinetic Energy of first body be T.

T=(MV2)/2

Where V is velocity of first body.

Let Kinetic Energy of second body be t.

t=(mv2)/2

Where v is the velocity of second body.

According to the given condition,

T=t

Therefore, (MV2)/2=(mv2)/2

Therefore, M/m=v2/V2

The ratio of their linear momenta would be,

P/p=MV/mv

where P and p is linear momentum of first and second body respectively.

Substituting for M/m, we get,

P/p=v/V

Thus, the ratio of linear momenta of two bodies of differing masses possesing equal kinetic energy is the reciprocal of the ratio of their velocities.

We can represent the ratio of their linear momenta in terms of the masses of the bodies using the ratio of the masses as obtaibed above.

M/m=v2/V2

Therefore, sqrt(M)/sqrt(m)=v/V

Substituting in the ratio of momenta,

P/p=sqrt(M)/sqrt(m)

Likewise, the ratio of the linear momenta of two bodies having equal Kinetic energy is the square root of the ratio of their masses

Answer 2

Hey there!

Given,

Two bodies m1 and m2

Let the velocity of 1st body be = v1

Then its Kinetic energy = p²/2m1=(m1v1)²/2m2

Let the velocity of 2nd body be = v2

Then its Kinetic energy = p²/2m2 = (m2v2)²/2m2

Now, (m1v1)²/2m1 = (m2v2)²/2m2

 ∴(m1v1)/(m2v2) = $\frac{ \sqrt{2m2} }{ \sqrt{2m1} }$

Then Ratio = √(2m2) : √(2m1)

Hope It Helps You!