Question

Two bodies have masses 2m and their Kinetic energies are in the ratio of 8:1 their momenta are in the ratio_?​

Answer 1

hey......... here is your answer.

given:- mass of first body = 2m.
mass of another body = m.
hence there kinetic energies are :- 1/2mv1^2÷1/2 mv2^2 =1/8
than,V1/V2=1/2.
than,the required ratio of momenta= 2m×2/m×1
= 4:1 (ans).

Answer 2

Correct QUESTION :-

Two bodies of masses 2 m and m have their K.E in the ratio 8:1. What is the ratio of their momenta ?

ANSWER :-

The mass of the bodies are given as 2 m and m .

We know that the kinetic energy of the body is given by :

K = 1/2 m v² .

For the first case :

Let the mass be 2 m and the velocity is v1 .

Let the Kinetic energy of the body be K1 .

Let the momentum be P1

For the second case :

The mass is m and the velocity is v1 .

Let the Kinetic energy of the body be K1 .

Let the momentum be P2 .

Deriving the kinetic energy and momentum :

K = 1/2 m v²

⇒ v² = 2 K/m ------(a)

Momentum = m v

⇒ p = m v

⇒ v = p/m

⇒ v² = p²/m² ---------(b)

From (a) and (b) we have :

p²/m² = 2 K/m

⇒ 2 K = p²/m

⇒ 2 K m = p²

⇒ p = √( 2 K m )

From this relation for the first case :

p1 = √( 2 K1 × m1 )

⇒ p1 = √( 2 × K1 × 2 m )

For the second case :

p2 = √( 2 × K2 × m )

⇒ p2 = √( 2 K2 m )

p1 : p2 = √( 4 K1 m ) : √( 2 K2 m )

Squaring both sides we get :

( p1 : p2 )² = ( 4 K1 m ) / ( 2 K2 m )

⇒ ( p1 : p2 )² = ( 2 K1 ) / ( K2 )

As per the question :

K1 / K2 = 8 / 1

⇒ K1 = 8 K2

⇒ ( p1 : p2 )² = ( 16 K2 ) / ( K2 )

⇒ ( p1 : p2 )² = 16

⇒ p1 : p2 = 4

⇒ p1 : p2 = 4 : 1

The ratio of momenta will be 4 : 1 .