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)
4 Answers

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
.