A 2.00-kg ball is moving at 2.20 m/s toward the right. It collides elastically with a 4.00-kg ball that is initially at rest. 1) Calculate the final velocity of the 2.00-kg ball. (Express your answer to three significant figures.)

Respuesta :

Elastic collision is when kinetic energy before = kinetic energy after

Ek= 1/2mv^2

total before
Ek=1/2(2)(2.2^2) = 4.84 J

total after
Ek= 1/2(2+4)(v^2) = 3v^2

Before = after
4.84=3v^2 | divide by 3
121/75 = v^2 | square root both sides
v=1.27 m/s

Answer:

The final velocity of the 2kg ball is 1.270 m/s

Explanation:

According to Newton's second and third laws of motion

Newton's second law state that "the rate of change of momentum is proportional to the applied force and takes place in the direction of that force".

Newton's third law state that "for every action, there must be an equal and opposite reaction".

The combinations of these two laws resulted in an elastic collision

Given that:

m1 = 2kg

u1 = 2.20m/s

m2 = 4.00kg

u2 = 0m/s

An Elastic collision is when kinetic energy before = kinetic energy after

E.K before = [tex]1/2mv^{2}[/tex]

E.K before = 1/2 * 2 * (2.20)^2

E.K = 1/2 * 2 * 4.84

E.K before = 4.84j

E.K after = 1/2 x (4 + 2)v^2

E.K after = 1/2(6v^2)

E.K after = 3v^2

Since E.K before = E.K after

4.84 = 3v^2

Divide through by 3

4.84/3 = 3v^2/3

1.6133 = v^2

[tex]V = \sqrt{1.6133} \\V = 1.270 m/s[/tex]