Question

(a) Calculate the initial momentum of each object and write the result in rectangular form.(b) Calculate the momentum of the combined mass after collision and write the result in polar form.(c) Calculate the final velocity after collision and write the result in polar form.

Answer 1

Given data:

Mass of ball is,

`m_b=5.75\text{ kg}`

Magnitude of initial velocity of ball is,

`v_(ib)=5.75\text{ m/s}`

Direction of initial velocity of ball with respect to x-axis is,

`\theta=30^o`

Magnitude of initial velocity of embeds is,

`v_(ie)=2.50\text{ m/s}`

Direction of initial velocity of embeds with respect to x-axis is,

`\theta_e=140^0`

Formula of momentum is as follows:

`\vec{P}=m\vec{v}\ldots(1)`

Initial momentum of ball,

Component of velocity of ball along x-axis is as follows:

`v_(ibx)=v_(ib)\cos \theta`

Substitute known values above equation,

`\begin{gathered} v_(ibx)=6.34*\cos 30^o \\ v_(ibx)=5.49\text{ m/s} \end{gathered}`

Component of velocity of ball along y-axis is as follows:

`\begin{gathered} v_(iby)=v\sin \theta \\ v_(iby)=6.34*\sin 30^o \\ v_(iby)=3.17\text{ m/s} \end{gathered}`

Therefore, momentum of ball is as follows:

`\begin{gathered} \vec{P}=m(v_(ibx)\hat{i}+v_(iby)\hat{j}) \\ \vec{P}=5.75\text{ kg(5.49}\hat{i}+3.17\hat{j}) \\ \vec{P}=(31.5675\hat{i}+18.2275\hat{j})kg\text{ }\cdot\text{ m/s} \end{gathered}`

Component of velocity of embeds along x-axis,

`\begin{gathered} v_(iex)=v_(ie)\cos \theta_e \\ v_(iex)=2.50*140^o \\ v_(iex)=-1.915\text{ m/s} \end{gathered}`

Component of velocity of embeds along y-axis,

`\begin{gathered} v_(iey)=v_(ie)\sin \theta_e \\ v_(iey)=2.50*\sin 140^o \\ v_(iey)=1.61\text{ m/s} \end{gathered}`

Initial momentum of embeds is as follows:

`\begin{gathered} \vec{P}=m_e(v_(iex)\hat{i}+v_(iey)\hat{j}) \\ \vec{P}=3.85\text{ kg(}-1.915\hat{i}+\hat{1.61\hat{j}})\text{ m/s} \\ \vec{P}=(-7.37\hat{i}+6.20\hat{j})\text{ kg}\cdot\text{ m/s} \end{gathered}`

Part-B:

After collision if both objects are stick together, this type of collision is called plastic collision.

Momentum of system remains constant before plastic collision and after plastic collision.

Hence,

`\vec{P}+\vec{P}_e=\vec{P}_f`
`\text{Here }\vec{\text{P}}_f\text{ is momentum of combine mass after collision}`

Substitute known values in above equation,

`\begin{gathered} \vec{P}_f=(31.5675\hat{i}+18.2275\hat{j})+(-7.37\hat{i}+6.20\hat{j})_{} \\ \vec{P}=(24.1975\hat{i}+24.4275\hat{j})\text{ kg}\cdot\text{ m/s} \end{gathered}`

Polar form of momentum of combined mass is as follows:

Magnitude of momentum of combined mass is as follows:

`\begin{gathered} P_f=\sqrt[]{(24.1975)^2+(24.4275)^2} \\ P_f=34.38\text{ kg}\cdot\text{ m/s} \end{gathered}`

Direction of momentum of combined mass,

`\begin{gathered} \theta_f=\tan ^(-1)(P_(fy))/(P_(fx)) \\ \theta_f=\tan ^(-1)((24.4275)/(24.1975)) \\ \theta_f=45^o \end{gathered}`

Polar form of momentum,

`P_f=(34.38\text{ kg}\cdot m/s,45^o)`

Part-C:

Final velocity of system is as follows:

Magnitude of final velocity is as follows:

`\begin{gathered} v=\sqrt[]{2.52^2+2.54^2} \\ v=3.58\text{ m/s} \end{gathered}`

Direction of final velocity with respect to x-axis,

`\begin{gathered} \theta_(fv)=\tan ^(-1)((2.54)/(2.52)) \\ \theta_(fv)=\tan ^(-1)(1.00) \\ \theta_(fv)=45^o \end{gathered}`

Polar form of final velocity,

`v=(3.58m/s,45^o)`