Answer:
a) P=0.8 Kg*m/s b) K=0.6 N c)P/K=MV/(1/2*MV²) d) V2f=1.5 m/s e) M2=0.53 Kg
Explanation:
During an elastic collision between 2 bodies, the momentum P is the same before and after the collision
For this case:
Before the collision:
M₁= mass of first car= 2 Kg
V₁= initial speed of the first car = 1 m/s
M₂= mass of the second car
V₂= initial speed of the second car = 0 m/s (as it is stationary)
After the collision:
V₁f= final speed of the first car after the collision= 0.6 m/s
V₂f= final speed of the second car after the collision
As momentum is the same after and before:
M₁V₁ + M₂V₂ = M₁V₁f + M₂V₂f consider that term M₂V₂=0 as V₂=0
Then, momentum for car N° 2 after the collision is: P₂= M₂V₂f and replacing from the above equation: P₂= M₁V₁ – M₁V₁f = M₁(V₁ – V₁f) = 2 Kg*(1m/s – 0.6m/s) = 0.8 Kg*m/s
As the kinetic energy “K” is also conservative:
½*M₁V₁² + ½*M₂V₂² = ½*M₁V₂f² + ½*M₂V₂f² Where ½*M₂V₂²=0
Then: K₂= ½*M₂V₂f² = ½*M₁(V₁² – V₁f²) = 0.64 N
Finally, to obtain M₂ and V₂f:
P₂=M₂V₂f and K₂=1/2*M₂V₂f2²
P₂/K₂= (M₂V₂f)/(1/2*M₂V₂f) =2/V₂f
V₂f= 2*K₂/P₂=1.5 m/s and M₂=P₂/V₂f=0.53 Kg
