Question

A player bounces a 0.48-kg soccer ball off her head, changing the velocity of the ball from vƒ— i = (8.1 m/s )x'' ( -2.1 m/s )y'' to vƒ— f = (5.6 m/s )x'' (4.0 m/s )y'' . What is the change in velocity of the ball?

1) 2.5 m/s
2) 3.5 m/s
3) 4.5 m/s
4) 5.5 m/s

Answer 1

To find the change in velocity of the soccer ball, subtract the initial velocity vector components from the final velocity vector components, and then calculate the magnitude using the Pythagorean theorem. The result is approximately 6.59 m/s, which isn't one of the provided options.

Step-by-step explanation:

The change in velocity of a soccer ball can be calculated using the initial and final velocity vectors given. The initial velocity, vi, is (8.1 m/s, -2.1 m/s), and the final velocity, vf, is (5.6 m/s, 4.0 m/s). To find the change in velocity, Δv, we subtract the initial velocity components from the final velocity components:

• Δvx = vfx - vix = 5.6 m/s - 8.1 m/s = -2.5 m/s
• Δvy = vfy - viy = 4.0 m/s - (-2.1 m/s) = 6.1 m/s

The magnitude of the change in velocity is then calculated using the Pythagorean theorem:

|Δv| = √((Δvx)2 + (Δvy)2) = √((-2.5 m/s)2 + (6.1 m/s)2) = √(6.25 + 37.21) m2/s2 = √43.46 m2/s2 ≈ 6.59 m/s

Therefore, the correct change in velocity of the ball is approximately 6.59 m/s which isn't listed in the provided options. It seems there might be a discrepancy or error in the options given.