91Ó°ÊÓ

We have given that the momenta${\left(p_x\right)}_a$to${\left(p_x\right)}_e$of thew object.

We need to find that the rank in order , from largest to smallest.

An object with mass $m$and moves with speed $v$having momentum $p$, this momentum is a vector and it is the product of the object mass and its velocity. A large mass of large velocity momentum. The momentum is given by equation $(11,3)$in the form

$p_x=mv\left(1\right)$

Object $\left(a\right)$:

The mass of the object is $m_a=20g$and velocity $v_a=1m/s$. We can use equation $\left(1\right)$and plug the values for $20g$and $1m/s$to get the momentum $p_a$

$$\begin{gathered} p_a=m_a{v}_a \\ =(20×{10}^{-3}kg)(1m/s) \\ =0.02kg(m/s) \end{gathered}$$

Object $\left(b\right)$:

The mass of object is $m_b=20g$and velocity $v_b=2m/s$. We can use equation $\left(1\right)$and putting the values for $20g$and $2m/s$to get momentum $p_b$

$$\begin{gathered} p_b=m_b{v}_b \\ =(20×{10}^{-3}kg)(2m/s) \\ =0.04kg(m/s) \end{gathered}$$

Object $\left(c\right)$:

Mass of the object is $m_c=10g$and velocity $v_c=2m/s$. We can use equation $\left(1\right)$and putting the values for $10g$and $2m/s$to get the momentum $p_C$

$$\begin{gathered} P_c=m_c{v}_c \\ =(10×{10}^{-3}kg)(2m/s) \\ =0.02kg(m/s) \end{gathered}$$

Object $\left(d\right)$:

Mass of the object is $m_d=10g$and velocity $v_d=1m/s$. We can use equation $\left(1\right)$and putting the values for$10g$and $1m/s$to get the momentum $p_d$

$$\begin{gathered} p_d=m_d{v}_d \\ =(10×{10}^{-3}kg)(1m/s) \\ =0.01kg(m/s) \end{gathered}$$

Object $\left(e\right)$:

Mass of the object is $m_e=200g$and velocity $v_e=0.1m/s$. We can use equation $\left(1\right)$and putting the values for $200g$and $0.1m/s$to get the momentum $p_e$

$$\begin{gathered} P_e=m_e{v}_e \\ =(200×{10}^{-3}kg)(0.1m/s) \\ =0.02kg(m/s) \end{gathered}$$

Finally, the order from the largest to the smallest is

$p_b>p_a=p_c=p_e>p_d$