As we know that two skateboarder will push each other so there is no external force on this system
Hence we can use momentum conservation here
[tex]m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}[/tex]
[tex] 0 + 0 = (65)(3.2) + 85(v)[/tex]
now we have
[tex]0 = 208 + 85 v[/tex]
[tex]v = - 2.45 m/s[/tex]
so the heavy skateboarder will move off with speed 2.45 m/s in the opposite direction
Conservation of linear momentum accounts for the direction of motion participating objects. The heavy skateboarder will move off with a speed of 2.45 m/s in the opposite direction.
As per the conservation of linear momentum, "When two objects approach each other, then the momentum before the collision is equal to the momentum after the collision.
The mass of the left skateboarder is m = 65 kg.
The mass of the right skateboarder is m' = 85 kg.
The speed of lighter skateboarder is, v = 3.2 m/s.
Applying the conservation of linear momentum as,
mu + m'u' = mv + m'v'
u and u' are the initial velocities of each skateboarder. Similarly, v and v' are their final velocities.
Solving as,
Since skateboarders are at rest initial, u = u' = 0.
m(0) + m'(0) = 65 × 3.2 + (85 × v)
v = -2.45 m/s
Thus, we can conclude that the heavy skateboarder will move off with a speed of 2.45 m/s in the opposite direction.
Learn more about the conservation of linear momentum here:
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