A 52 kg woman is standing in a 26 kg cart that is moving a velocity of 1.2 m/s S. The woman staying rest then moves at a velocity of 1m/s (E) what is the final velocity of the cart?

Question

I just ugh, don't even know how to even start answering the question as this deals with momentum and I ve been used to working with collision problems. South and east are set as + directions and cart is m1 ,, woman is m2. I know that momentum is conserved so Pi equals to pf but no clue how to go further than this

Thank you

Answer 1

Using Earth as the frame of reference:
Note: c = cart and p=person/woman

conservation of momentum in y-dir:
m_p * v_py + m_c * v_cy = m_p * v_py' + m_c * v_cy'
note: the first term of left m_p * v_py' is 0 because the person is moving west relative to ground (i.e. no motion in the y-dir). Now, you need to isolate for v_cy'
v_cy' = (m_p * v_py + m_c * v_cy) / m_c
= (52 * 1.2 + 26 * 1.2) / 26
= 3.6 m/s [S] --> the cart is still moves to S but with higher magnitude

conservation of momentum in the x-dir:
m_p * v_px + m_c * v_cx = m_p * v_px' + m_c * v_cx'
Note: all terms on the left side are 0 because the cart and person were moving in the y-dir before (i.e. there is no velocity in x. So, there is no momentum in the x)
0 = m_p * v_px' + m_c * v_cx'
isolate for v_cx'
v_cx' = (m_p * v_px') / m_c
= (52 * 1) / 26
= 2 m/s [E]

Now recombine the vectors v_cy' and v_cx' in order to find the final velocity of cart after this interaction. *use pythagorean theorem for magnitude and tan for angle*