The mass of the baseball is $m = 0.142 mathrm{~kg}$. The speed of the baseball is $v = 45.1 mathrm{~m/s}$.

$textbf{Required: }$

Finding the magnitude of the momentum of the baseball.

In order to evaluate the magnitude of the Momentum of the baseball, we use the formula that is given in the second yellow box:

$$
begin{align*}
p &= m ~ v \
&= 0.142 mathrm{~kg} times 45.1 mathrm{~m/s} \
&= 6.4042 mathrm{~kg cdot m/s} \
end{align*}
$$

So, the magnitude of the Momentum of the baseball is $6.4042 mathrm{~kg cdot m/s}$.

$$
textbf{Solution:}
$$

The formula for momentum is.

$$
p=mv
$$

indent Using the above definition we can say that

$$
Rightarrow v = frac{p}{m}=dfrac{37;kgcdot m/s}{12;kg}=boxed{bf 3.1m/s}
$$

Then momentum for each object can find by product of mass and speed for each of the case:

$textbf{Solution:}$

$$
p_A=m_A v_A=(10,kg)(10,m/s)=100,kg.m/s
$$

$$
p_B=m_B v_B=(15,kg)(4,m/s)=60,kg.m/s
$$

$$
p_C=m_C m_C=(5,kg)(20,m/s)=100,kg.m/s
$$

$$
p_D=m_D m_D=(60,kg)(3,m/s)=180,kg.m/s
$$

In order to arrive at result, we compare the results of momentum:

$$
boxed{bf color{#4257b2}p_B< p_A=p_D<p_C}
$$