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
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]