Chapter 7, Problem 34P

To determine

The kinetic energy acquired by each piece

Answer to Problem 34P

  $4491\text{J and }3009\text{J}$

Explanation of Solution

Given:

Mass of the lighter piece $=m_1=m$

Mass of the heavier piece $=m_2=1.5m$

Final velocity of lighter piece after explosion $=v_{1i}$

Final velocity of heavier piece after explosion $=v_{2i}$

Initial total momentum of the object $=p_i=0$

Final total momentum of the object $=p_f$

Total kinetic energy after the explosion $=K{E}_f=7500\text{J}$

Kinetic energy of lighter piece after the explosion $=K{E}_1$

Kinetic energy of heavier piece after the explosion $=K{E}_2$

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, we have

  $\begin{array}{l}{p}_i=p_f\\ \text{Where}\\ p_i=\text{initial total momentum }\\ p_f=\text{final total momentum}\end{array}$

Kinetic energy is given

  $KE=(0.5)m{v}^2$

Calculation:

According to conservation of momentum,

  $\begin{array}{l}{p}_i=p_f\\ 0=p_f\\ 0=m_1{v}_{1i}+m_2{v}_{2i}\\ 0=m{v}_{1i}+(1.5)m{v}_{2i}\\ v_{2i}=\frac{-2v_{1i}}{3}\left(\text{1}\right)\end{array}$

Total kinetic energy after the explosion is given as

  $\begin{array}{l}K{E}_f=(0.5)m_1{v}_{1i}^2+(0.5)m_2{v}_{2i}^2\\ \end{array}$

Using equation (1)

  $\begin{array}{l}K{E}_f=(0.5)m_1{v}_{1i}^2+(0.5)m_2{v}_{2i}^2\\ 7500=(0.5)m{v}_{1i}^2+(0.5)(1.5)m{\left(\frac{-2v_{1i}}{3}\right)}^2\\ 7500=(0.5)m{v}_{1i}^2+(0.5)(1.5)m{\left(\frac{-2v_{1i}}{3}\right)}^2\\ 7500=(1.67)(0.5)m{v}_{1i}^2\\ (0.5)m{v}_{1i}^2=4491\\ K{E}_1=4491\text{J}\end{array}$

Total kinetic energy after the explosion is given as

  $\begin{array}{l}K{E}_f=K{E}_1+K{E}_2\\ 7500=4491+K{E}_2\\ K{E}_2=3009\text{J}\end{array}$

Conclusion:

Therefore, kinetic energy of the pieces comes out to be $4491\text{J and }3009\text{J}$