Acart with mass 2.0 kg moving on a frictionless linear air track at an initial speed of 1.0 m/s undergoes an elastic collision with an initially stationary cart of unknown mass m. after the collision, the first cart continues in its original direction at 0.6 m/s. let v be the velocity of the second cart after the collision. a) what is the momentum p = mv of the second cart after the collision? b) what is the kinetic energy k = 1 2mv2 of the second cart after the collision? c) divide the equation from (a). what is the velocity v of the second cart after the collision? (b) by the equation from d) what is the mass m of the second cart?

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

8 meters above the spring