Chapter 9.2, Problem 23PP

To determine

The velocity of the rocket after the fuel has burned.

Answer to Problem 23PP

  $7.91\text{m/s}$

Explanation of Solution

Given:

The mass of the model rocket, $m_r=4\text{kg}$

The mass of the fuel in the rocket, $m_F=50\text{g = 0}\text{.0}5\text{kg}$

Speed at which fuel is burned, $v_{ff}=625\text{m/s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

And, law of conservation of linear momentum: Momentum is conserved in a closed and isolated system and is given by,

  $p_i=p_f$

Calculation:

Assume the downward direction as negative and upward direction as positive. Then, the velocity of the burning fuel will be written as,

  $v_{ff}=-625\text{m/s}$

Consider the rocket and fuel as a system and ignore air resistance, then it will be a closed and isolated system and law of conservation of momentum can be applied to it.

Initially the rocket-fuel combined system was rest, thus

  $v_f=0\text{m/s}$

Let, the velocity of the rocket after the fuel has burned $v_{rf}$ , then using the law of conservation of linear momentum,

  $\begin{array}{c}{p}_i=p_f\\ \left(m_r+m_f\right)v_i=\left(m_r-m_f\right)v_{rf}+m_f{v}_{ff}\end{array}$

Using given values in above,

  $\begin{array}{c}{p}_i=p_f\\ \left(4+0.05\right)\left(0\right)=\left(4-0.05\right)v_{rf}+\left(0.05\right)\left(-625\right)\\ v_{rf}=7.91\text{m/s}\end{array}$

Positive sign shows the upward direction of the rocket’s motion.

Conclusion:

Thus, the velocity of the rocket after the fuel has burned is $7.91\text{m/s}$ .