A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the moving puck is 6.0 m/s. After the collision, one puck leaves with a speed v1 at 30° to the original line of motion. The second puck leaves with speed v2 at 60°. Calculate v1 and v2.

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]

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