Question

A 600-g mass traveling at 8.0 m/s undergoes a head-on elastic collision with a 200-g mass traveling toward it also at 8.0 m/s. What is the speed of the 200-g mass after the collision?

Answer 1

Answer:

vf₂ = 16 m/s

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 = 600 g

m₂= 200 g

v₀₁ = 8 m/s :initial velocity of m₁  

v₀₂ = - 8 m/s :initial velocity  of m₂  ; (-) indicates that m₂ is moving in the opposite direction to m₁  

Problem development

We appy the formula (1):

We assume that at the end of the collide both masses move in the initial direction of m₁ , then vf₁ and vf₁ have a sign (+)

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂

(600) *(8)+ (200)*( -8) = (600) *vf₁ +  (200)*vf₂  

4800 - 1600 = (600) *vf₁ +  (200)*vf₂  

3200 =  (600) *vf₁ +  (200)*vf₂  :  We divide both sides of the equation by 100

32= 6*vf₁ +2*vf₂ Equation (1)

For elastic collision, the elastic restitution coefficient (e) is equal to 1 :

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

$1= \frac{ v_{f2}- v_{f1} }{ 8- (-8) }$

16 = vf₂  - vf₁

vf₁ = vf₂  - 16  Equation (2)

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

32= 6*vf₁ +2*vf₂

32= 6*( vf₂  - 16) +2*vf₂

32= 6*vf₂  - 96 + 2*vf₂

32 + 96= 6*vf₂ + 2*vf₂

128 = 8 *vf₂

vf₂ = 128 / 8

vf₂ = 16 m/s