Final answer:
The impulse delivered by the wall to the ball, when considering a collision that reverses the x-component of the ball's velocity without changing its magnitude, is given by '10 * m (Ns)', where 'm' is the mass of the ball. None of the given options (A-D) match this calculation.
Step-by-step explanation:
To determine the impulse delivered by the wall to a ball, we must consider the change in the ball's momentum. Impulse is defined as the change in momentum of an object when a force is applied over a period of time. The momentum p of an object is given by the product of its mass m and its velocity v, that is, p = m * v. For the given problem, the ball's speed stays the same, but the direction of its velocity changes before and after hitting the wall.
Initially, the ball moves at an angle of 60° above the +x-direction, which means that its x-component of velocity is vx = v * cos(60°). The final velocity of the ball is at the same angle, but now above the -x-direction, meaning the x-component of velocity will be -vx(since only the direction has changed and not the magnitude). Therefore, the change in the x-component of the velocity is Δvx = -vx - vx = -2 * vx or Δvx = -2 * v * cos(60°). With the mass m of the ball, the change in momentum or impulse J is given by J = m * Δvx.
For a ball with a mass m and initial speed v = 10 m/s, the impulse delivered by the wall, considering only the x-component, will be
J = m * (-2 * 10 * cos(60°))
Now, because cos(60°) = 0.5, we have
J = m * (-2 * 10 * 0.5) = -10 * m (Ns).
Thus, the impulse in terms of m, the mass of the ball, would be -10 * m (Ns), which is not presented in the options A to D of the question as stated.