Answer:
a) [tex]\sf \sqrt{\dfrac{5}{8}}v [/tex]
b) [tex]\bold{\sf \dfrac{3}{4} m v^2}[/tex]
Explanation:
Let's go through the given information and solve for the velocity of the coalesced mass [tex]\bold{\sf (V)}[/tex] and then find the loss of energy during the collision.
Given information:
1. Law of conservation of linear momentum in the X-direction:
- [tex]\sf mv = 4mV \cos(\alpha) [/tex] ...........(i)
2. Law of conservation of linear momentum in the Y-direction:
- [tex]\sf 3mv = 4mV \sin(\alpha) [/tex] ...........(ii)
3. Find [tex]\bold{\sf \tan(\alpha)}[/tex]:
- [tex]\sf \tan(\alpha) = 3 [/tex] ...........(iii)
Now, let's find [tex]\bold{\sf \cos(\alpha)}[/tex] and [tex]\bold{\sf \sin(\alpha)}[/tex] using the given information:
- [tex]\sf \cos(\alpha) = \dfrac{v}{4V} [/tex] ...........(iv)
- [tex]\sf \sin(\alpha) = \dfrac{3v}{4V} [/tex] ...........(v)
Now, use [tex]\bold{\sf \cos^2(\alpha) + \sin^2(\alpha) = 1}[/tex] to find [tex]\bold{\sf V}[/tex]:
- [tex]\sf \left(\dfrac{v}{4V}\right)^2 + \left(\dfrac{3v}{4V}\right)^2 = 1 [/tex] ...........(vi)
Now, solve for [tex]\bold{\sf V}[/tex]:
[tex]\sf \dfrac{v^2}{16V^2} + \dfrac{9v^2}{16V^2} = 1 [/tex]
[tex]\sf \dfrac{10v^2}{16V^2} = 1 [/tex]
[tex]\sf V^2 = \dfrac{10v^2}{16} [/tex]
[tex]\sf V= \sqrt{\dfrac{10v^2}{16}} [/tex]
[tex]\sf V = \sqrt{\dfrac{5}{8}}v [/tex]
So, the velocity of the coalesced mass [tex]\bold{\sf (V)}[/tex] is:
[tex]\sf \sqrt{\dfrac{5}{8}}v [/tex]
[tex]\begin{aligned} \cline \end{aligned}[/tex]
b) Loss of Energy During the Collision:
The loss of energy ([tex]\bold{\Delta KE}[/tex]) is the initial kinetic energy minus the final kinetic energy:
[tex]\sf \Delta KE = KE_{\textsf{initial}} - KE_{\textsf{final}} [/tex]
The initial kinetic energy is the sum of the kinetic energies of particles A and B:
[tex]\sf KE_{\textsf{initial}} = \dfrac{1}{2} m v^2 + \dfrac{1}{2} (3m) v^2 [/tex]
The final kinetic energy is the kinetic energy of the coalesced mass:
[tex]\sf KE_{\textsf{final}} = \dfrac{1}{2} (4m)V^2 [/tex]
Substitute the values of [tex]\bold{\sf V}[/tex] into the expressions for initial and final kinetic energies:
[tex]\sf KE_{\textsf{initial}} = \dfrac{1}{2} m v^2 + \dfrac{1}{2} (3m) v^2 [/tex]
[tex]\sf KE_{\textsf{final}} = \dfrac{1}{2} (4m) \left(\sqrt{\dfrac{4}{5}}v\right)^2 [/tex]
Calculate [tex]\bold{\sf \Delta KE}[/tex]:
[tex]\sf \Delta KE = \dfrac{1}{2} m v^2 + \dfrac{1}{2} (3m) v^2 - \dfrac{1}{2} (4m) \left(\sqrt{\dfrac{5}{8}}v\right)^2 [/tex]
[tex]\sf \Delta KE = \dfrac{1}{2} m v^2 + \dfrac{1}{2} (3m) v^2 - \dfrac{1}{2} (4m) \left(\dfrac{5}{8}\right) v^2 [/tex]
[tex]\sf \Delta KE = \dfrac{1}{2} m v^2 + \dfrac{1}{2} (3m) v^2 - 2\cdot \dfrac{5}{8} m v^2 [/tex]
[tex]\sf \Delta KE = \dfrac{1}{2} m v^2 + \dfrac{3}{2} mv^2 -\dfrac{5}{4} m v^2 [/tex]
[tex]\sf \Delta KE = mv^2 \left( \dfrac{1}{2} + \dfrac{3}{2} -\dfrac{5}{4} \right) [/tex]
[tex]\sf \Delta KE = mv^2 \left( \dfrac{ 1\cdot 2 + 3\cdot 2 - 5 }{4} \right)[/tex]
[tex]\sf \Delta KE = \left(\dfrac{2+6-5}{4} \right)m v^2 [/tex]
[tex]\sf \Delta KE = \dfrac{3}{4} m v^2 [/tex]
Therefore, the loss of energy during the collision is:
[tex]\bold{\sf \dfrac{3}{4} m v^2}[/tex]
