Engineering Mechanics: Statics & Dynamics (14th Edition)

$1.6m/s$
We can determine the required velocity as follows: We apply the law of conservation of momentum in the x-direction: $m_c\cdot v_{c, 1x}+m_p\cdot v_{p,1x}=(m_c+m_p)v_x$ This simplifies to: $v_x=\frac{m_p\cdot v_{p,1x}}{m_c+m_p}$.eq(1) As $\frac{v_{p,1}}{v_{p,1x}}=\frac{5}{4}$ $\implies v_{p,1x}=\frac{4v_{p,1}}{5}=8m/s$ We plug in the known values in eq(1) to obtain: $v_x=\frac{5.8}{20+5}=1.6m/s$