Final answer:
The impulse delivered by the wall to the ball with mass 'm' and speed 10 m/s moving at a 60° angle is -5m N·s in the -x-direction. It results from the change in only the horizontal component of velocity upon impact with the wall.
Step-by-step explanation:
The question relates to the concept of impulse in physics. Impulse is defined as the change in momentum of an object when a force is applied over a time period, which is expressed by the equation J = Δp = m(vf - vi), where J is the impulse, Δp is the change in momentum, m is the mass, vf is the final velocity, and vi is the initial velocity.
For the scenario where a ball of mass m traveling at 10 m/s at an angle of 60° with respect to the +x-direction bounces off a wall and moves at the same speed but at a 60° angle above the -x-direction, we can calculate the impulse. Considering only the horizontal component of the velocity since the vertical component does not change upon bouncing off a vertical wall, the initial and final velocities in the x-direction are vi_x = v*cos(60°) and vf_x = -v*cos(60°), respectively. Therefore, the change in momentum (Δp_x) is m * -2v*cos(60°). Since the magnitude of cosine 60° is 0.5, the impulse J delivered by the wall to the ball is J = -m * 2 * 10 m/s * 0.5 = -5m N·s. The direction of this impulse is in the -x-direction.