Chapter 7, Problem 63GP

To determine

The components of final velocity of the rocket.

Answer to Problem 63GP

The x component of velocity is $v_x{}^{\prime }=\frac{3}{2}{v}_0$ .

The y component of velocity is $v_y{}^{\prime }=-v_0$ .

Explanation of Solution

Given:

The mass of the rocket is $m$ .

The mass of the fuel dumped is $\frac{m}{3}$ .

The velocity of the rocket along x direction is $v_0$ .

The velocity of the rocket along y direction is $v_y=\frac{1}{3}m=2v_0$ .

Formula used:

The expression for momentum is,

  $p=mv$

Here, m is the mass and v is the velocity.

The expression for law of conservation of momentum is,

  $m_1{u}_1=m_1{v}_1$

Here, $m_1$ is the mass of the object 1, $u_1$ is the initial velocity of object 1, and $v_1$ is the final velocity of object 1.

Calculation:

Sketch the representation of the rocket as shown below.

  

Apply the momentum along x direction.

  $\begin{array}{l}{m}_1{v}_0=\left(m_1-m_2\right)v_x{}^{\prime }+m_2{v}_1\\ m{v}_0=\left(m-\frac{1}{3}m\right)v_x{}^{\prime }\\ v_x{}^{\prime }=\frac{3}{2}{v}_0\end{array}$

Apply the momentum along y direction.

  $\begin{array}{l}0=m_2{v}_y+\left(m_1-m_2\right)v_y{}^{\prime }\\ 0=\frac{1}{3}m\left(2v_0\right)+\left(m-\frac{1}{3}m\right)v_y{}^{\prime }\\ -\frac{2}{3}m{v}_0=\frac{2}{3}m{v}_y{}^{\prime }\\ v_y{}^{\prime }=-v_0\end{array}$

Conclusion:

Thus, the x component of velocity is $v_x{}^{\prime }=\frac{3}{2}{v}_0$ and the y component of velocity is $v_y{}^{\prime }=-v_0$ .