Take into account that the total momentum of the system before andafter the clay sticks to the car, must be the same.
Then, you can write:
mv1 + Mv2 = (m + M)v
where,
m: mass of the clay = 0.16kg
M: mass of the toy car = 0.51kg
v1: initial speed of the clay = 5 m/s
v2: initial speed of the car = 0 m/s
v: speed of both clay and car after the impact = ?
Solve the equation above for v, replace the values of the other parameters and simplify:
![\begin{gathered} v=(mv_1+Mv_2)/(m+M)=\frac{(0.16kg)(5(m)/(s))+(0.51kg)(0(m)/(s))}{0.16\operatorname{kg}+0.51\operatorname{kg}} \\ v\approx1.2(m)/(s) \end{gathered}]()
Hence, the speed of both clay and toy car after the impact is approximately 1.2m/s