Given:
The mass of the pucks, m=0.300 kg
The initial velocity of the 1st puck, u₁=3.33 m/s
The initial velocity of the 2nd puck, u₂=1.87 m/s
The final velocity of the 1st puck, v₁=2.10 m/s
To find:
The y-component of the final velocity of the 2nd puck.
Explanation:
Let us assume that the north is the positive y-axis and the east is the positive x-axis.
As the 1st puck is going northwest, the angle made by the velocity of the 1st puck with ponegative-axis is 45 °
The velocities can be represented in the vector form as,
[tex]\begin{gathered} \vec{u_1}=3.33cos45\degree(-\hat{i})+3.33sin45\degree\hat{j} \\ =-2.35\hat{i}+2.35\hat{j} \end{gathered}[/tex]And
[tex]\vec{u_2}=1.87(-\hat{j})=-1.87\hat{j}[/tex]And
[tex]\vec{v_1}=-2.10\hat{i}[/tex]Where i-cap and j-cap are the unit vectors along the positive x and y axes respectively.
From the law of conservation of momentum, the total momentum of a system remains the same at all times. The momentum is conserved simultaneously and independently along both axes.
Considering the y-axis,
[tex]\begin{gathered} mu_1sin45\degree\hat{j}+mu_2(-\hat{j})=0+mv_{2y}\hat{j} \\ \Rightarrow u_1sin45\degree-u_2=v_{2y} \end{gathered}[/tex]Where v_2y is the y-component of the final velocity of the second puck.
On substituting the known values,
[tex]\begin{gathered} 3.33sin45\degree-1.87=v_{2y} \\ \Rightarrow v_{2y}=0.48\text{ m/s} \end{gathered}[/tex]Final answer:
The y-component of the final velocity of the second puck is 0.48 m/s
