Aman of mass 80kg stands next to a stationary ball of mass 4kg on a frictionless surface. he kicks the ball forward along the surface with a speed 15m/s. calculate the man’s recoil speed.

We cause conservation of momentum.  The initial momentum of the situation is zero, since momentum is given by p=mv.  Initially v of the ball and man are zero so the total momenta (the sum of the ball's and man's momenta) is zero.
Conservation of momentum means the vector sum of the final momenta must equal zero.  That is, they will be equal and opposite.  When the ball goes forward at 15m/s the momentum is p=(4kg)x(15m/s)=60kgm/s.
The man must have the same momentum, but in the opposite direction (negative).  We can find his speed by considering the magnitude of his velocity:


Checking our work we see that an 80kg man moving at 0.75m/s gives
(80kg)(0.75m/s)=60kgm/s, as expected.

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