Asmall ball of mass 2.00 kilograms is moving at a velocity
1.50 meters/second. it hits a larger, stationary ball of mass 5.00 kilograms. what is the kinetic energy of the system after the collision if the collision is elastic?

Kf= 2.24 J : Kinetic energy of the system after the collision .

Explanation:

Theory of collisions

Linear momentum is a vector magnitude (same direction and direction as velocity) and its magnitude is calculated like this:

P = m*v

where

p : Linear momentum

m : mass

v : velocity

There are 3 cases of collisions : elastic, inelastic and plastic.

For the three cases the total linear momentum quantity is conserved:

P₀=Pf  Formula (1)

P₀ :Initial  linear momentum quantity

Pf : Initial  linear momentum quantity

Nomenclature and data

m₁: small ball  mass= 2kg

V₀₁: initial big ball  speed, = 1.5 m/s

Vf₁: final small ball  speed

m₂: big ball mass = 5 kg  

V₀₂: initial big ball speed, = 0

Vf₂: final big ball speed

Problem development

For this problem the collision is perfectly elastic ,then, In addition to the linear moment, the knetic energy is also conserved.

We assume that the small ball moves to the right before the collision and continues moving  to the right after the collision .(+)

We assume that the big ball moves to the right before the collision.(+)

We apply formula (1)

P₀=Pf

m₁*V₀₁+m₂*V₀₂=m₁*Vf₁+m₂*Vf₂

(2)*(1.5(+(5)*(0) =(2)*(Vf₁)+(5)*(Vf₂)

3 = 2*Vf₁+5*Vf₂   Equation (1)

For perfectly elastic collision the coefficient of elastic restitution (e) is equal to 1, and e is defined like this:

1*(V₀₁-V₀₂) =Vf₂-Vf₁  , V₀₂=0, V₀₁ =1.5 m/s

1.5=Vf₂-Vf₁

Vf₁=Vf₂-1.5 Equation (2)  

We replace Equation (2) in Equation (1)

3= 2(Vf₂-1.5)+5*Vf₂

3= 2*Vf₂-3+5*Vf₂

6=7Vf₂

Vf₂ =6/7 = 0.857 m/s

We replace Vf₂ = 0.857 m/s  in the  Equation (2)  

Vf₁=0.857 -1.5 = -0.643 m/s : (-) the small ball moves in the opposite direction  of the movement before the collision.

Kinetic energy of the system after the collision if the collision (Kf)

Kf=(1/2) m₁*Vf₁²+ (1/2)*m₂*Vf₂²

Kf=(1/2)( 2)*(-0.643)²+ (1/2)*(5)*(0.857)= 2.24 J

two charges are fixed in place on the x-axis as shown. the charge at x = 0 is negative and has magnitude q = 2 μcoulombs (1 μcoulomb = 10-6 coulombs). the charge at x = d, (d = +12 cm), is positive and has magnitude q = 4 μcoulombs. find a point x = xo along the x-axis at which the electric field is zero.

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