Question

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.

Answer 1

• 
  `v_1  =  \ 5.196 (m)/(s)`

• 
  `v_2 =  3 (m)/(s)`

Step-by-step explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

`\vec{p}_i = \vec{p}_f`

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

`\vec{p}_i = m_1 \ \vec{v}_(1_i) + m_2 \ \vec{v}_(2_i)`

as the second puck is initially at rest:

`\vec{v}_(2_i) = 0`

Using the unit vector
`\vec{i}` pointing in the original line of motion:

`\vec{v}_(1_i) = 6.0 (m)/(s) \hat{i}`

`\vec{p}_i = 0.70 \ kg  \ 6.0 (m)/(s) \ \hat{i} + 0.70 \ kg \ 0`

`\vec{p}_i = 4.2 \ (kg \ m)/(s) \ \hat{i}`

So:

`\vec{p}_i =  4.2 \ (kg \ m)/(s) \ \hat{i} = \vec{p}_f`

`\vec{p}_f =  4.2 \ (kg \ m)/(s) \ \hat{i}`

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

`\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )`

So, our velocity vectors will be:

`\vec{v}_(1_f) = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )`

`\vec{v}_(2_f) = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )`

We got

`\vec{p}_f = 0.7 \ kg \ \vec{v}_(1_f) + 0.7 \ kg \ \vec{v}_(2_f)`

`4.2 \ (kg \ m)/(s) \ \hat{i} = 0.7 \ kg \   v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )  + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )`

So, we got the equations:

`4.2 \ (kg \ m)/(s)  = 0.7 \ kg \   v_1 \  cos(30 \°) + 0.7 \ kg \ v_2 \  cos(-60 \°)`

and

`0  = 0.7 \ kg \   v_1 \  sin(30 \°) + 0.7 \ kg \ v_2 \  sin(-60 \°)`.

From the last one, we get:

`0  = 0.7 \ kg \  ( v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°) )`

`0  =  v_1 \  sin(30 \°) +  \ v_2 \  sin(-60 \°)`

`v_1 \  sin(30 \°) = -  \ v_2 \  sin(-60 \°)`

`v_1  =  \ v_2 \  (sin(60 \°))/( sin(30 \°) )`

and, for the first one:

`4.2 \ (kg \ m)/(s)  = 0.7 \ kg  \ (  v_1 \  cos(30 \°) + v_2 \  cos(60 \°) )`

`(4.2 \ (kg \ m)/(s))/( 0.7 \ kg) =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)`

`(4.2 \ (kg \ m)/(s))/( 0.7 \ kg) =    v_1 \  cos(30 \°) + v_2 \  cos(60 \°)`

`6 \ (m)/(s) =    (\ v_2 \  (sin(60 \°))/( sin(30 \°) ) ) \  cos(30 \°) + v_2 \  cos(60 \°)`

`6 \ (m)/(s) = v_2     (\   (sin(60 \°))/( sin(30 \°) ) ) \  cos(30 \°) +   cos(60 \°)`

`6 \ (m)/(s) = v_2  * 2`

so:

`v_2 = 6 \ (m)/(s) / 2 = 3 (m)/(s)`

and

`v_1  =  \ 3 (m)/(s)  \  (sin(60 \°))/( sin(30 \°) )`

`v_1  =  \ 5.196 (m)/(s)`