contestada

A woman of mass 57 kg jumps off the bow of
a 42 kg canoe that is intially at rest.
If her velocity is 1.5 m/s to the right, what
is the velocity of the canoe after she jumps?

Respuesta :

Here we will use the momentum conservation

Since there is no external force on this system so momentum of woman + canoe system will always remains conserved

So here initial momentum of the system is zero as they are at rest so final momentum will always be zero

[tex]m_1 v_1 + m_2v_2 = 0[/tex]

[tex]57(1.5) + 42(v) = 0[/tex]

[tex]85.5 + 42v = 0[/tex]

[tex]v = -\frac{85.5}{42}[/tex]

[tex]v = - 2.03 m/s[/tex]

so the canoe will go back with speed 2.03 m/s

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