Final answer:
The impulse delivered by the wall to a ball bouncing off it with the same speed but in the opposite direction is determined by the change in momentum. With an initial velocity of 10 m/s angled at 60°, the impulse can be calculated by considering the initial and final momenta in the x-direction. The impulse magnitude is -10m N·s, where m is the mass of the ball.
Step-by-step explanation:
The impulse delivered by the wall to the ball can be computed by applying the principles of conservation of momentum and understanding the change in momentum due to the collision. Since the ball reverses its horizontal direction while maintaining the same speed upon bouncing off the wall, the change in momentum in the x-direction can be calculated. To find the impulse, we consider the initial momentum and the final momentum of the ball. The impulse is equal to the change in momentum, which is the difference between the final and initial momentum vectors.
The initial velocity in the x-direction is 10 m/s \(\times\) cos(60°), and after the bounce its velocity in the x-direction is -10 m/s \(\times\) cos(60°). Given that momentum p = m \(\times\) v, we multiply the mass m of the ball by these velocities to find the initial and final momentum. Therefore, the impulse J, delivered by the wall, can be expressed as:
J = m \(\times\) (final velocity in the x-direction - initial velocity in the x-direction)
By plugging in the known values we get:
J = m \(\times\) ([-10 \(\times\) cos(60°)] - [10 \(\times\) cos(60°)]) = m \(\times\) (-5 - 5) = -10m N\(\cdot\)s
The negative sign indicates that the impulse is directed in the negative x-direction.