A 11 g object moving to the right at 29 cm/s overtakes and collides? elastically with a 24 g object moving in the same direction at 19 cm/s. Find the velocity of the slower object after the collision. AND find the velocity of the faster object after the collision. Answer in units of cm/s.

Respuesta :

Answer:

vf₁  =  15.29 cm/s : to the right

vf₂ =  25.29 cm/s : to the right

Explanation:

Theory of collisions  

Linear momentum is a vector magnitude (same direction of the 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 : Final linear momentum quantity  

Data

m₁= 11 g : mass of  object₁

m₂= 24 g  : mass of  object₂

v₀₁ = 29 cm/s , to the right : initial velocity of m₁

v₀₂=  19 cm/s, to the right  i :initial velocity of m₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

We assume that the two objects move to the right at the end of the collision, so, the sign of the final speeds is positive:

( 11)*( 29) + (24 )*(19) = ( 11)*vf₁ +(24)*vf₂

775 = ( 11)*vf₁ +(24)*vf₂ Equation (1)

Because the shock is elastic, the coefficient of elastic restitution (e) is equal to 1.

[tex]e = \frac{v_{f2} -v_{f1}}{v_{o1} -v_{o2}}[/tex]

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(29 -  19 )  = (vf₂ -vf₁)

10 =  (vf₂ -vf₁)

vf₂ = 10 + vf₁  Equation (2)

We replace Equation (2) in the Equation (1)

775 = ( 11)*vf₁ +(24)*(10 + vf₁ )

775 = ( 11)*vf₁ +240+(24) vf₁

775 - 240= ( 35)*vf₁

535 = ( 35)*vf₁

vf₁  = 535 / 35

vf₁  =  15.29 cm/s : to the right : Final velocity of object₁  

We replace vf₁  =  15.29  cm/s in the Equation (2)

vf₂ = 10 + vf₁

vf₂ =10 +  15.29

vf₂ =  25.29 cm/s, to the right: Final velocity of object₂