# ratio of their masses?

An object at rest is suddenly broken apart into two fragments by an explosion. The first fragment (with mass m_1) acquires twice the kinetic energy of the second one (with mass m_2).

What is the ratio of their masses? (m_1/m_2)

• by conservation of momentum m1 v1 = m2 v2 , where v1 and v2 are the velocities of the fragments after explosion. then, v1 / v2 = m2 / m1

now, as per question, K1 = 2 K2, where K1 andK2 are their kinetic energies.hence, 1/2 m1 v1^2 = 2 x 1/2 m2 v2^2

ie., 2x m2 / m1 = (v1 /v2)^2= (m2 / m1)^2

then, m1 / m2 = 1 / 2 ie., m1 : m2 = 1 : 2 is the required ratio.

• Using the relation KE = P² / 2m where P is the momentum and KE the kinetic energy.

m_1 has momentum P_1 and kinetic energy P²_1 / 2m_1

m_2 has momentum P_2 and kinetic energy P²_2 / 2m_2

By conservation of momentum, |P_1| = |P_2| = P (say)

We are told KE of m_1 is twice KE of m_2

so P² / 2m_1 = 2 P² / 2m_2

Multiplying through by 2 / P²

1 / m_1 = 2 / m_2

From which 2m_1 = m_2

2m_1 / m_2 = 1

m_1 / m_2 = 1/2

Ratio of mass 1 to mass 2 is 1 : 2

• nicely the respond is A yet that's not the style you clean up it. Q=3 A=2 Q/A ratio is .574 so then enable us to divide Q by using A that's 3/2=a million.5 Now multiply your answer by using the ratio. a million.5*.574=.861

• The momentums have to be equal, so we have:

m₁V₁ = m₂V₂

From the KE statement

2*½m₁V₁² = ½m₂V₂²

2m₁V₁² = m₂V₂²

now we have 2 equations

m₁V₁ = m₂V₂

2m₁V₁² = m₂V₂²

m₁/m₂ = V₂/V₁

from the other equation

2m₁V₁² = m₂V₂²

V₁² / V₂² = m₂/2m₁

V₁ / V₂ = √(m₂/2m₁)

substituting

m₁/m₂ = √(2m₁/m₂)

squaring

m₁²/m₂² = 2m₁/m₂

multiplying both sides by m₂/m₁

m₁/m₂ = 2

.