Question

A woman of mass 57 kg jumps off the bow of

Answer 1

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

$m_1 v_1 + m_2v_2 = 0$

$57(1.5) + 42(v) = 0$

$85.5 + 42v = 0$

$v = -\frac{85.5}{42}$

$v = - 2.03 m/s$

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