A giant rubber ball of mass 3500 kg and velocity 45.0 m s^-1, collides elastical

Question

A giant rubber ball of mass 3500 kg and velocity 45.0 m s^-1, collides elastically and head on with a smaller rubber ball of mass 200 kg and velocity 22.0 ms^-1 moving in the opposite direction. The diagram is not drawn to scale. (a) What is an elastic collision and what do we know about the relative motion of each ball before and after the collision? (b) Find the speed of separation of the two balls after collision. (c) Hence or otherwise, calculate the final velocities of the two rubber balls.

Explanation / Answer

(a) Always remember one thing. Momentum shall be always conserved whether the collision is elastic or inelastic.

Howvere, in the case of elstic collision, one additional quantity KINETIC ENERGY shall also be conserved.

So, the answer of your first part is elastic collison is that type of collision in which momentum as well as the kinetic energy of the colliding bodies remain conserved.

Again, in the elastic collision relative speed of each ball shall be equal before and after the collision.

(b) Consider the velocity of the two ball are v1 and v2 after the collision.

apply conservation of momentum -

3500*45 - 200*22 = 3500v1 + 200v2

=> 3500v1 + 200v2 =153100

=> 17.5v1 + v2 = 765.5

=> v2 = 765.5 - 17.5v1---------------------------------------------------(i)

now apply conservation of energy -

(1/2)*3500*45^2 + (1/2)*200*22^2 = (1/2)*3500*v1^2 + (1/2)*200*v2^2

=> 3500*v1^2 + 200*v2^2 = 7184300

=> 17.5*v1^2 + v2^2 = 35921.5

put the value of v2 from (i) -

=> 17.5*v1^2 + (765.5 - 17.5v1)^2 = 35921.5

=> 323.7*v1^2 - 26792.5*v1 + 550068.75 = 0

=> v1^2 - 82.8*v1 + 1699 = 0

=> v1 = [82.8 + sqrt(6855.84 - 6796)] / 2 = [82.8 + 7.7]/2 = 45.25 m/s

the other value of v1 is -

[82.8 - sqrt(6855.84 - 6796)] / 2 = [82.8 -7.7]/2 = 37.6 m/s

The first value of v1 is greater than 45 m/s so it can be discarded.

So, we have v1 = 37.6 m/s

and v2 = 765.5 - 17.5*37.6 = 107.5 m/s

So the speed of seoaration of two ball after the collision = v2 - v1 = 107.5 - 37.6 = 69.9 m/s.

(c) Final velocity of the rubber balls -

v1 = 37.6 m/s

v2 = 107.5 m/s.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.