Question

An angry rhino with a mass of 2600 kg charges directly toward you with a speed of 3.70 m/s. Before you start running, as a distraction, you throw a 0.180-kg rubber ball directly at the rhino with a speed of 8.39 m/s. Determine the speed of the ball after it bounces back elastically toward you.

Answer 1

Answer:

vf₂ = 15.79 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 = 2600 kg : mass of the rhino

m₂= 0.18 kg : mass of the ball

v₀₁ = 3.70 m/s  : initial velocity of the rhino

v₀₂= - 8.39 m/s, initial velocity of the ball

Problem development

We appy the formula (1):

P₀ = Pf  

m₁*v₀₁ + m₂*v₀₂ = m₁*vf₁ + m₂*vf₂  

(2600)*(3.7) + (0.18)*(- 8.39) = (2600)*vf₁ +(0.18)*vf₂

9620 -1.5102 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)*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)

( 3.7 -( -8.39 )  = (vf₂ -vf₁)

12.09 =  (vf₂ -vf₁)

vf₂ = vf₁ + 12.09 Equation (2)

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

9618.4898 = (2600)*vf₁ +(0.18)*vf₂

9618.4898 = (2600)*vf₁ +(0.18)* (vf₁ + 12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + (0.18)(12.09)

9618.4898 = (2600)*vf₁ +(0.18)*vf₁ + 2.1762

9618.4898 -2.1762 = (2600.18)*vf₁

9616.3136 = (2600.18)*vf₁  

vf₁ = (9616.3136) / (2600.18)

vf₁ = 3.698 m/s : Final velocity of the rhino

We replace vf₁ = 3.698 m/s in the Equation (2)

vf₂ = vf₁ + 12.09

vf₂ = 3.698 + 12.09

vf₂ = 15.79 m/s : Final velocity of the ball