91影视

From the figure we can observe that,

Mass of the object A, $m_A=5.0kg$

Velocity of object A,$v_A=v_{Ay}=-11\mathrm{m}/\mathrm{s}$

Mass of the object B,$m_B=6.0\mathrm{kg}$

Velocity and angle of the object B, ${\mathrm{v}}_{\mathrm{B}}=10.0\mathrm{m}/\mathrm{s},\mathrm{胃}=60掳$

Mass of the object C,$m_C=10\mathrm{kg}$

Velocity of the object C,$v_C=v_{Cx}=-3.0m/s$$v_C=v_{Cx}=-3.0\mathrm{m}/\mathrm{s}$

A property of a moving body resulting from its mass and motion that is roughly equivalent to the product of its mass and velocity: a property of a moving body that determines how long it will take for it to come to rest when applied to a constant force.

As we know the momentum of an object is

$\stackrel{鈫�}{p}=m\stackrel{鈫�}{v}$

If $p_x$is x component of momentum and $p_y$is y component of momentum

Then

$p_x+p_y=m{\stackrel{鈫�}{v}}_x+m{\stackrel{鈫�}{v}}_y$

Therefore,

The x-component of the net momentum of the system consists of A and C can be given as.

$$\begin{gathered} p_{Ax}+p_{Cx}=m_A{v}_{Ax}+m_c{v}_{cx} \\ =50\mathrm{kg}\left(0\right)+10\mathrm{kg}-\left(-3\mathrm{m}/\mathrm{s}\right) \\ =-30\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The x component of the net momentum of the system consists of A and C is -30kg.m/s.

The y component of the net momentum of the system consists of A and C can be given as.

$$\begin{gathered} p_{Ax}+p_{Cx}=m_A{v}_{Ay}+m_c{v}_{cy} \\ =5\mathrm{kg}\left(-11\mathrm{m}/\mathrm{s}\right)+10\mathrm{kg}\left(0\right) \\ =-55.0\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The y component of the initial momentum of the system consists of A and C is -55.0kg/m/s

Given in the question.

Mass of the object B,$m_B=6.0\mathrm{kg}$

Velocity and angle of the object B,$v_B=10.0\mathrm{m}/\mathrm{s},\mathrm{胃}=60掳$

Therefore,

X-component of the velocity of object B, $v_{Bx}=10.0\cos 60\mathrm{m}/\mathrm{s}$

Y-component of velocity of the object B,$v_{By}=10.0\sin 60\mathrm{m}/\mathrm{s}$

Mass of object C,$m_C=10kg$

Velocity of object$v_C=v_{cx}=-3.0\mathrm{m}/\mathrm{s}$

The x component of the net momentum of the system consists of B and C can be given as.

$$\begin{gathered} p_{Bx}+p_{Cx}=m_B{v}_{Bx}+m_c{v}_{cx} \\ =6\mathrm{kg}\left(10\cos 60\mathrm{m}/\mathrm{s}\right)+10\mathrm{kg}\left(-3\mathrm{m}/\mathrm{s}\right) \\ =30\mathrm{kg}.\mathrm{m}/\mathrm{s}-30\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =0 \end{gathered}$$

The x component of the net momentum of the system consists of B and C is 0.

The y component of the net momentum of the system consists of B and C can be given as.

$$\begin{gathered} p_{By}+p_{Cy}=m_B{v}_{By}+m_c{v}_{cy} \\ =6\mathrm{kg}\left(10\cos 60\mathrm{m}/\mathrm{s}\right)+10\mathrm{kg}\left(0\right) \\ =52\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The y component of the net momentum of the system consists of B and C is 52kg.m/s .

We know,

$$\begin{gathered} {\mathrm{m}}_{\mathrm{A}}=5.0\mathrm{kg} \\ {\mathrm{v}}_{\mathrm{A}}={\mathrm{v}}_{\mathrm{Ay}}=-11\mathrm{m}/\mathrm{s} \\ {\mathrm{m}}_{\mathrm{B}}=6.0\mathrm{kg} \\ {\mathrm{v}}_{\mathrm{Bx}}=10.0\cos 60\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{By}}=10.0\sin 60\mathrm{m}/\mathrm{s} \\ {\mathrm{m}}_{\mathrm{C}}=10\mathrm{kg} \\ {\mathrm{v}}_{\mathrm{C}}={\mathrm{v}}_{\mathrm{Cx}}=-3.0\mathrm{m}/\mathrm{s} \end{gathered}$$

The x component of the net momentum of the system consists of A, B and C can be given as.

$$\begin{gathered} p_{Ax}+p_{Bx}+p_{Cx}=m_A{v}_{ax}+m_B{v}_{Bx} \\ =5\mathrm{kg}\left(0\right)+6\mathrm{kg}\left(10\cos 60\mathrm{m}/\mathrm{s}\right)+10\mathrm{kg}\left(-3\mathrm{m}/\mathrm{s}\right) \\ =30\mathrm{kg}.\mathrm{m}/\mathrm{s}-30\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =0 \end{gathered}$$

The x component of the net momentum of the system consists of A, B and C is 0 .

The y component of the net momentum of the system consists of A, B and C can be given as.

$$\begin{gathered} p_{Ay}+p_{By}+p_{Cy}=m_A{v}_{ay}+m_B{v}_{By} \\ =5\mathrm{kg}\left(-11\mathrm{m}/\mathrm{s}\right)+6\mathrm{kg}\left(10\sin 60\mathrm{m}/\mathrm{s}\right)+10\mathrm{kg}\left(0\right) \\ =\left(-55+52\right)\mathrm{kg}.\mathrm{m}/\mathrm{s} \\ =-3\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$

The y component of the net momentum of the system consists of A, B and C is -3kg.m/s .