Question

A. What is the total momentum of the system before the collision? (Write your answer in terms of the mass, m)B. What is the total momentum of the system after the collision? (Write your answer in terms of the mass, m)C. Write puck 1’s velocity after the collision in component form. D. What is the y-component of puck 2’s velocity after the collision?E. What is the x-component of puck 2’s velocity after the collision?F. At what angle does puck 2 move after the collision? Determine the angle and draw it on a diagram. G. What is the magnitude of puck 2’s velocity after the collision?

Answer 1

Part (A)

The total momentum of the system before collision is given as,

`p_(ix)=m_1u_(1x)+m_2u_(2x)`

Here, m_1 is the mass puck 1, m_2 is the mass of puck 2, u_1x is the x-component of initial velocity of puck 1 (u_1x=-10 m/s), and u_2x is the x-component of initial velocity of puck 2 (u_2x=8 m/s).

Substituting all known values,

`\begin{gathered} p_(ix)=m*(-10\text{ m/s})+m*(8\text{ m/s}) \\ =-2m\text{ kg}.\text{ m/s} \end{gathered}`

As, both puck does not have any initial velocity component along y-direction. The momentum along y-direction will be zero.

Therefore, the total momentum of the system before collision is,

`\begin{gathered} p_i=p_(ix)\hat{i}+p_(iy)\hat{j} \\ =-2m\hat{i}\text{ kg}.\text{ m/s} \end{gathered}`

Part (b)

According to the law of conservation of momentum, the momentum before and after the collision remains constant. Therefore, the total momentum after collision will be,

`\begin{gathered} p_f=p_i \\ =-2m\hat{i}\text{ kg}.\text{ m/s} \end{gathered}`

Part (C)

The x-component of the velocity of puck 1 after collision is given as,

`v_(1x)=v_1\cos \theta`

Here, v_1 is the velocity of puck 1 after collision (v_1=6 m/s), and θ is the angle made with x-axis (θ=25°).

Substituting all known values,

`\begin{gathered} v_(1x)=(6\text{ m/s})*\cos (25^(\circ)) \\ =5.44\text{ m/s} \end{gathered}`

The y-componet of velocity of puck 1 after collision is given as,

`v_(1y)=v_1\sin \theta`

Substituting all known values,

`\begin{gathered} v_(1y)=(6\text{ m/s})*\sin (25^(\circ)) \\ =2.54\text{ m/s} \end{gathered}`

Therefore, the velocity of puck 1 after collision in vector form is given as,

`\begin{gathered} v_1=v_(1x)\hat{i}+v_(1y)\hat{j} \\ =(5.44\hat{i}+2.54\hat{j})\text{ m/s} \end{gathered}`

Part (D)

Applying law of conservation of momentum along y-direction.

`m_1u_(1y)+m_2u_(2y)=m_1v_(1y)+m_2v_(2y)`

Here, v_2y is the velocity of puck 2 along y direction after collision.

Substituting all known values,

`\begin{gathered} m*0+m*0=m*(2.54\text{ m/s})+m* v_(2y) \\ 0=2.54\text{ m/s}+v_(2y) \\ v_(2y)=-2.54\text{ m/s} \end{gathered}`

Part (E)

Applying law of conservation of momentum along x-direction.

`m_1u_(1x)+m_2u_(2x)=m_1v_(1x)+m_2v_(2x)`

Here, v_2x is the velocity of puck 2 along x direction after collision.

Substituting all known values,

`\begin{gathered} m*(-10\text{ m/s})+m(8\text{ m/s})=m*(5.44\text{ m/s})+mv_(2x) \\ -2m=5.44m+mv_(2x) \\ -2=5.44+v_(2x) \\ v_(2x)=-2-5.44 \\ v_(2x)=-7.44\text{ m/s} \end{gathered}`

Part (F)

The angle at which puck 2 move after collision is given as,

`\begin{gathered} \tan \phi=(v_(2y))/(v_(2x)) \\ \phi=\tan ^(-1)((v_(2y))/(v_(2x))) \end{gathered}`

Substituting all known values,

`\begin{gathered} \phi=\tan ^(-1)(\frac{-2.54\text{ m/s}}{-7.44\text{ m/s}}) \\ =18.85^(\circ) \end{gathered}`

Part (G)

The magnitude of puck 2's velocity after the collision is given as,

`v_2=\sqrt[]{v^2_(2x)+v^2_(2y)}`

Substituting all known values,

`\begin{gathered} v_2=\sqrt[]{(-7.44\text{ m/s})^2+(-2.54\text{ m/s})^2} \\ =7.86\text{ m/s} \end{gathered}`