What is the answer to this problem

Use conservation of momentum. (or Newtons first law)
and also use conservation of mass

Assume elastic collision

m1v1 + m2v2 = m1V1 + m2V2
if we move things over and group like terms

m1(v1 - V1) = m2(V2 - v2)

m1 = 4kg
m2 = 1kg
v1 = 6.5m/s
v2 = 0
V1 = 4.8
V2 = ?

m1(v1 - V1) = m2(V2 - v2)
4(6.5-4.8) = 1(x-0)

(expand and group)

x = 6.8m/s

(Other ways to do this!)

It equals just about 6,300 km per hour each day an week

Answer: a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same sign as the  -4.0  µc charge.======electricstatic force between two point charges  is given by    where k is coulomb's constant,  8.99×10⁹  n·m²/c², q₁ and q₂ are the point charges, and r is the distance between the two point charges.assign positive to q₁let q₁ =  6.0 µc, q₂ = -4.0  µc, q₀ represent the charge at the origin, and q₃ represent this new charge to place.the charge at the origin q₀ would experience the following force f₀₁ from q₁, with r = 6.0 m (note that 1  µc = 10⁻⁶ c)    the force from q₂    the force from q₃    the net force on the origin charge, q₀, has to be zero for no electrostatic forcehaving an equation where k and q₀ can be divided out of the equation will allow us to solve for q₃    isolating:     a charge of -1.0  × 10⁻⁴ c must be placed  at x = 26 m.the sign of the charge will be the same as the  -4.0  µc charge