Respuesta :
Answer:
Al's mass is 102.92 kg
Explanation:
As there are no external forces in the horizontal direction, the horizontal net force must be zero:
[tex]F_{net} = 0[/tex]
As the force is the derivative in time of the momentum, this means that the horizontal momentum is constant:
[tex]F_{net} = \frac{dp_{horizontal}}{dt} = 0[/tex]
[tex]p_{horizontal_i }= p_{horizontal_f}[/tex]
where the suffix i and f means initial and final respectively.
The initial momentum will be:
[tex]p_{horizontal_}i = m_{Al} \ v_{Al_i} + m_{Jo} \ v_{Jo_i}[/tex]
But, as they are at rest, initially
[tex]p_{horizontal_i} = m_{Al} * 0 + m_{Jo} * 0[/tex]
[tex]p_{horizontal_i} = 0[/tex]
So, this means:
[tex]p_{horizontal_f} = m_{Al} \ v_{Al_f} + m_{Jo} \ v_{Jo_f} = 0 [/tex]
We know that the have an combined mass of 195 kg:
[tex]m_{total} = m_{Al} + m_{Jo} = 195 \ kg[/tex].
so:
[tex] m_{Jo} = 195 \ kg - m_{Al}[/tex].
[tex]m_{Al} \ v_{Al_f} + (195 \ kg - m_{Al}) \ v_{Jo_f} = 0 [/tex]
[tex]m_{Al} \ v_{Al_f} - m_{Al} \ v_{Jo_f}= - 195 \ kg \ v_{Jo_f} [/tex]
[tex]m_{Al} \ (v_{Al_f} - v_{Jo_f})= - 195 \ kg \ v_{Jo_f} [/tex]
[tex]m_{Al} = \frac{ - 195 \ kg \ v_{Jo_f} } { v_{Al_f} - v_{Jo_f} } [/tex]
[tex]m_{Al} = \frac{195 \ kg \ v_{Jo_f} } { v_{Jo_f} - v_{Al_f} } [/tex]
Now, we can use the values:
[tex]v_{Al_f}= 10.2 \frac{m}{s}[/tex]
[tex]v_{Jo_f}= - 11.4 \frac{m}{s}[/tex]
where the minus sign appears as they are moving at opposite directions
[tex]m_{Al} = \frac{195 \ kg ( - 11.4 \frac{m}{s} ) } { (- 11.4 \frac{m}{s}) - 10.2 \frac{m}{s} } [/tex]
[tex]m_{Al} = 102.92 \ kg [/tex]
and this is the Al's mass.