Question

A woman of mass 49 kg jumps off the bow of

a 46 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?
Answer in units of m/sˆı.

Answer 1

`v_c_a_n_o_e=-1.6m/s^(2)`

Step-by-step explanation:

You can use the equation of the conservation of momentum. We find that the velocity of the canoe is -1.6m/s².

`m_1v_1+m_2v_2=0\\(49)(1.5)+(46)(v_c_a_n_o_e)=0\\73.5 + 46v_c_a_n_o_e=0\\46v_c_a_n_o_e = -73.5\\v_c_a_n_o_e = -73.5/46 = -1.6m/s^(2) \\`