Suppose a watermelon with a mass of 2.0 kg undergoes a head-on elastic collision on a frictionless counter with a grapefruit with a mass of 0.8 kg. If the total kinetic energy of the system is 10.5 J and the total momentum is 7.5 kg·m/s, determine the possible initial and final velocities for the watermelon and the grapefruit.  

 

Answer:

for an elastic collision both the kinetic energy and the momentum are conserved.

let V1 be velocity of watermelon and V2 the velocity of grapefruit.

thereforefor kinetic energy,

-1/2m₁v1² +1/2m2v₂² = 10.5

(0.5 *2* v1^2) +( 0.5*0.8*v2^2) =10.5

v1^2 + 0.4v2^2 =10.5

10v1^2 +4v2^2 = 105..................(i)

momentum equation;

m₁u₁ +m₂u₂ = 7.5

2v1 +0.8 v2 =7.5

20v1 +8v2 =75......................(ii)

solving equation 1 and 2 simultaneously gives;

from equation 2,

v2 = (75-20v1)/8 .............................(iii)

substitute equation 3 in eqn 1 to get;

10v1^2 +4{(75-20v1)/8}² =105

this gives;

V1 = 1.606m/s or -0.1639m/s

to get the velocity 2, we substitute the v1 in either equation to get;

v₂ = (75-20*1.606)/8 =5.36m/s

or V2 = (75-20*-0.1639)/8 =9.785m/s

therefore,

V₁ =1.606m/s when V₂ = 5.36m/s

or

V₁ = -0.1639m/s when V₂ =9.785m/s

Step-by-step explanation

we use the laws of conservation of momentum and kinetic energy to find the energy before and energy after collision as well as the momentum before and after collision. this gives a simultaneous equation which is solved to obtain the required values of velocity of melon and grapefruit.