Answer:
Kf= 2.24 J : Kinetic energy of the system after the collision .
Step-by-step 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