Chapter 7, Problem 74GP

To determine

The ratio of masses

Answer to Problem 74GP

  $1:2$

Explanation of Solution

Given:

Mass of the object before explosion $=m$

Velocity of object before explosion $=V=0$

Mass of the first fragment $=m_1$

Mass of the second fragment $=m_2$

Velocity of first fragment after explosion $=v_{1f}$

Velocity of second fragment after explosion $=v_{2f}$

Kinetic energy of second fragment is twice the kinetic energy of first fragment.

Formula Used:

Momentum of an object is given as

  $\begin{array}{l}p=mv\\ \text{where }p=\text{momentum}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

According conservation of momentum principle, momentum of the system before explosion is same as the momentum after explosion.

Also, kinetic energy is given as

  $KE=(0.5)m{v}^2$

Calculation:

According to conservation of momentum for the explosion of the object,

  $\begin{array}{l}\text{Momentum of first fragment + momentum of second fragment = momentum of the object}\\ m_1{v}_{1f}+m_2{v}_{2f}=mV\\ m_1{v}_{1f}+m_2{v}_{2f}=m(0)\\ m_1{v}_{1f}+m_2{v}_{2f}=0\\ v_{1f}=\frac{-m_2{v}_{2f}}{m_1}\left(1\right)\end{array}$

Also it is given that

  $\begin{array}{l}\text{kinetic energy of second fragment = 2 (kinetic energy of first fragment)}\\ (0.5)m_1{v}_{1f}{}^2=(2)(0.5)m_2{v}_{2f}{}^2\\ m_1{v}_{1f}{}^2=2m_2{v}_{2f}{}^2\\ \text{Using equation }\left(1\right)\\ m_1{\left(\frac{-m_2{v}_{2f}}{m_1}\right)}^2=2m_2{v}_{2f}{}^2\\ \left(\frac{m_1{m}_2}{m^2{}_1}\right)=2\\ \frac{m_1}{m_2}=\frac{1}{2}\end{array}$

Conclusion:

Therefore, the ratio is $1:2$