Answer
The velocity of the third part is $~~(1000~\hat{i}-166.7~\hat{j})~m/s$
Work Step by Step
We can find the initial momentum of the body:
$p_i = (20.0~kg)(200~m/s) = 4000~kg~m/s$
We can write an expression for the initial momentum in unit-vector form:
$p_i = (4000~\hat{i})~kg~m/s$
We can express the momentum of the first part in unit-vector form:
$p_1 = (10.0~kg)(100~m/s)~\hat{j} = (1000~\hat{j})~kg~m/s$
We can express the momentum of the second part in unit-vector form:
$p_2 = (4.00~kg)(-500~m/s)~\hat{i} = (-2000~\hat{i})~kg~m/s$
We can use conservation of momentum to find the momentum of the third part in unit-vector form:
$p_1+p_2+p_3 = p_i$
$p_3 = p_i-p_1-p_2$
$p_3 = (4000~kg~m/s)~\hat{i}-(1000~kg~m/s)~\hat{j}-(-2000~kg~m/s)~\hat{i}$
$p_3 = (6000~\hat{i}-1000~\hat{j})~kg~m/s$
Note that the mass of the third part must be $6.00~kg$
We can express the velocity of the third part in unit-vector form:
$m_3~v_3 = p_3$
$v_3 = \frac{p_3}{m_3}$
$v_3 = \frac{(6000~kg~m/s)~\hat{i}-(1000~kg~m/s)~\hat{j}}{6.0~kg}$
$v_3 = (1000~\hat{i}-166.7~\hat{j})~m/s$
The velocity of the third part is $~~(1000~\hat{i}-166.7~\hat{j})~m/s$
