Answer:
a) [tex]v_{3}=\frac{V\sqrt{2}}{3}[/tex]
b) [tex]v_{3}=11.79 m/s[/tex]
c) [tex]\alpha=45^{\circ}[/tex]
Explanation:
a) Let's use the conservation of momentum, it means the change of momentum in the system is equal to zero.
The inial total momentum is zero and the final total momentum is the sum of each particle momentum.
[tex]0=m_{1}v_{1}+m_{2}v_{2}+m_{3}v_{3}[/tex]
But we know that [tex]m_{1}=m_{2[/tex] and [tex]v_{1}=v_{2}=V[/tex] and [tex]m_{3}=3m_{1}[/tex], so we have the vectorial equation:
[tex]0=m_{1}V\hat{i}+m_{1}V\hat{j}+3m_{1}\vec{v_{3}}[/tex]
Now we can find the vector v₃ in terms of V.
[tex]\vec{v_{3}}=-\frac{V}{3}(\hat{i}^{2}+\hat{j}^{2}) [/tex]
Let's find now the magnitude of this vector:
[tex]v_{3}=\frac{V}{3}\sqrt{(\hat{-i})^{2}+(\hat{-j})^{2}}=\frac{V\sqrt{2}}{3}[/tex]
b) We just need to replace V in the above equation;
[tex]v_{3}=\frac{25\sqrt{2}}{3}=11.79 m/s [/tex]
c) The vector velocity of the particle 3 is in term of components, x, and y. So we can use tangent to find the angle.
[tex]tan(\alpha)=\frac{V_{x}}{V_{y}}[/tex]
but we know that Vx and Vy are the same so we have:
[tex]tan(\alpha)=1[/tex]
[tex]\alpha=arctan(1)=45^{\circ}[/tex]
I hope it helps you!
