Question

A 10.9 g object moving to the right at 18.5 cm/s makes an elastic head-on collision with a 17.1 g object moving in the opposite direction at 29.4 cm/s. What is the velocity of the 10.9 g object after the collision (assume positive to the right)?

Answer 1

Answer:

vf₁ = - 40 cm/s ; velocity of the 10.9 g object after the collision

vf₁ =  40 cm/s ,  to the left

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 = 10.9 g : mass of the object₁

m₂ = 17.1 g : mass of the object₂

v₀₁ =  18.5 cm/s , to the right : initial velocity of the object₁

v₀₂=  29.4 cm/s, to the left  :initial velocity of the object₂

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

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

( 10.9 )*(18.5) - ( 17.1)*(29.4) = ( 10.9 )*vf₁ + ( 17.1)*vf₂

-301.09 = (10.9)*vf₁ +(17.1)*vf₂ Equation (1)

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

$e= \frac{v_{f2}-v_{f1} }{v_{o1}-v_{o2}}$

1*(v₀₁ - v₀₂ )  = (vf₂ -vf₁)

(18.5 -( -29.4) )  = (vf₂ -vf₁)

47.9 = vf₂ -vf₁

vf₂ =  47.9 +vf₁ Equation (2)

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

-301.09 = (10.9)*vf₁ +(17.1)*vf₂

-301.09 = (10.9)*vf₁ +(17.1)*(47.9 +vf₁)

-301.09 = (10.9)*vf₁ + 819.09 +17.1vf₁

-301.09-819.09 = (10.9)vf₁ + (17.1 )vf₁

-1120.18 = 28vf₁

vf₁ = -1120.18 / 28

vf₁ = - 40 cm/s ; velocity of the 10.9 g object after the collision

vf₁ =  40 cm/s ,  to the left