# A cardinal (Richmondena cardinalis) of mass 4.20 times 10^{-2} kg and a baseball of mass 0.150 kg...

## Question:

A cardinal (Richmondena cardinalis) of mass {eq}4.20 \times 10^{-2}\ kg {/eq} and a baseball of mass {eq}0.150\ kg {/eq} have the same kinetic energy. What is the ratio of the cardinal's magnitude {eq}p_c {/eq} of momentum to the magnitude {eq}p_b {/eq} of the baseball's momentum?

## Kinetic Energy:

The kinetic energies of objects are the component of their total energies that correspond to their movement. We can express the kinetic energy in terms of the mass, m, and the velocity, v, wherein {eq}\displaystyle KE = \frac{1}{2}mv^2 {/eq}.

We first express the kinetic energies of the moving bodies using the equation, {eq}\displaystyle KE = \frac{1}{2}mv^2 {/eq}. We have the mass, m, and the velocity, v of the object. With this expression, we, then, find the velocity, v.

We have the expression for the kinetic energy of the cardinal, {eq}\displaystyle KE_C {/eq}.

{eq}\begin{align} \displaystyle KE_C &= \frac{1}{2}m_C v^2_C\\ \frac{2KE_C}{m_C} &= v_C^2\\ \sqrt{\frac{2KE_C}{m_C}} &= v_C \end{align} {/eq}

Now, we consider the expression for the kinetic energy of the baseball, {eq}\displaystyle KE_B {/eq}.

{eq}\begin{align} \displaystyle KE_B &= \frac{1}{2}m_B v^2_B\\ \frac{2KE_B}{m_B} &= v_B^2\\ \sqrt{\frac{2KE_B}{m_B}} &= v_B \end{align} {/eq}

Now, we take the ratio of the momentum, {eq}\displaystyle p = mv {/eq}, of the given objects. Afterwards, we simplify the equation using the given values:

• {eq}\displaystyle m_C = 4.20\times 10^{-2}\ kg {/eq}
• {eq}\displaystyle m_{B} = 0.150\ kg {/eq}
• {eq}\displaystyle KE_C = KE_B {/eq}

We proceed with the solution.

{eq}\begin{align} \displaystyle \frac{p_C}{p_B} &= \frac{m_Cv_C}{m_Bv_B}\\ &= \frac{m_C\left( \sqrt{\frac{2KE_C}{m_C}} \right)}{m_B\left( \sqrt{\frac{2KE_B}{m_B}} \right)}\\ &= \frac{\sqrt{m_C}}{\sqrt{m_B}}\\ &= \sqrt{ \frac{4.20\times 10^{-2}\ kg}{0.150\ kg}}\\ &\approx\boxed{\rm 0.529} \end{align} {/eq}