Question

An object at rest is suddenly broken apart into two fragments by an explosion. One fragment acquires twice the kinetic energy of the other. What is the ratio of their masses?

Short answer

The ratio of masses of the fragments A and B is \(1:2\).

Step-by-step solution

Conservation of Momentum

According to the law of conservation of linear momentum, the momentum of a closed, isolated system remains conserved when no external force acts on the system.

For example, consider a system of two objects that collide head-on with each other. Suppose there is no external force acting on the system.

In that case, the total momentum of the system before the collision is equal to the total momentum after the collision, i.e., the total momentum of the system remains conserved.

Given information

The object is at rest initially, so the initial velocity of the object is\(u = 0\;{\rm{m/s}}\)

Object explodes into two fragments: A and B.

Let velocities of the fragments A and B be\({v_1}\)and\({v_2}\), respectively, and masses of fragments A and B are\({m_1}\)and\({m_2}\), respectively.

Thus, the mass of the object before the explosion is \(m = {m_1} + {m_2}\).

Application of the law of conservation of the momentum

The momentum of the object before the explosion is

\(\begin{array}{c}{p_{\rm{i}}} = \left( {{m_1} + {m_2}} \right)u\\ = 0\end{array}\).

The momentum of the object after the explosion is\({p_f} = {m_1}{v_1} + {m_2}{v_2}\).

According to the law of conservation of momentum, the initial momentum of the object is equal to the final momentum of the object, i.e.,

\(\begin{array}{c}{p_{\rm{i}}} = {p_{\rm{f}}}\\0 = {m_1}{v_1} + {m_2}{v_2}\\{m_1}{v_1} = {m_2}{v_2}\\{v_1} = \frac{{{m_2}{v_2}}}{{{m_1}}}\end{array}\) ... (i)

Determination of the ratio of masses of the fragments

Let\(K{E_1}\)and\(K{E_2}\)be the kinetic energies of fragments A and B, respectively. If the kinetic energy of fragment 1 is twice the kinetic energy of fragment 2, then

\(\begin{array}{c}K{E_1} = 2K{E_2}\\\frac{1}{2}{m_1}v_1^2 = 2 \times \frac{1}{2}{m_2}v_2^2\\\frac{1}{2}{m_1}v_1^2 = {m_2}v_2^2\end{array}\)

On substituting the value of\({v_1}\)from equation (i) in the above expression,

\(\begin{array}{c}\frac{1}{2}{m_1}{\left( {\frac{{{m_2}{v_2}}}{{{m_1}}}} \right)^2} = {m_2}v_2^2\\\frac{{{m_2}}}{{{m_1}}} = \frac{2}{1}\\\frac{{{m_1}}}{{{m_2}}} = \frac{1}{2}\end{array}\)

Thus, you can say that the fragment with higher kinetic energy has a lower mass, and the ratio of masses of fragments A and B is \(1:2\).