Respuesta :
Answer:
- [tex] v_1 = \ 5.196 \frac{m}{s} [/tex]
- [tex] v_2 = 3 \frac{m}{s} [/tex]
Explanation:
For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:
[tex]\vec{p}_i = \vec{p}_f[/tex]
where the suffix i means initial, and the suffix f means final.
The initial momentum will be:
[tex]\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}[/tex]
as the second puck is initially at rest:
[tex]\vec{v}_{2_i} = 0[/tex]
Using the unit vector [tex]\vec{i}[/tex] pointing in the original line of motion:
[tex]\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}[/tex]
[tex]\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0[/tex]
[tex]\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}[/tex]
So:
[tex]\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f[/tex]
[tex]\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i} [/tex]
Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula
[tex]\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )[/tex]
So, our velocity vectors will be:
[tex] \vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )[/tex]
[tex] \vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )[/tex]
We got
[tex]\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}[/tex]
[tex] 4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )[/tex]
So, we got the equations:
[tex] 4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)[/tex]
and
[tex] 0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)[/tex].
From the last one, we get:
[tex] 0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )[/tex]
[tex] 0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) [/tex]
[tex] v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°) [/tex]
[tex] v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } [/tex]
and, for the first one:
[tex] 4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )[/tex]
[tex] \frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°) [/tex]
[tex] \frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°) [/tex]
[tex] 6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°) [/tex]
[tex] 6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°) [/tex]
[tex] 6 \ \frac{m}{s} = v_2 * 2 [/tex]
so:
[tex] v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s} [/tex]
and
[tex] v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) } [/tex]
[tex] v_1 = \ 5.196 \frac{m}{s} [/tex]