A tennis ball of mass 7.5×10−2 kgkg and speed 21 m/sm/s strikes a wall at a 45 ∘∘ angle and rebounds with the same speed at 45 ∘∘.(Figure 1)Figure1 of 1Part AWhat is the magnitude of the impulse given to the ball?

The impulse given to an object is equal to the change in its momentum. The momentum of an object is given by the product of its mass and its velocity.

In this case, the tennis ball's initial momentum is its mass times its initial velocity. The direction of the momentum is the same as the direction of the velocity, so it's at a 45 degree angle to the wall.

The final momentum of the ball is its mass times its final velocity, but in the opposite direction (since it rebounded off the wall).

The change in momentum, or impulse, is the final momentum minus the initial momentum.

Let's calculate:

1. Initial momentum (p_initial) = mass * velocity = 7.5 * 10^-2 kg * 21 m/s = 1.575 kg*m/s
2. Final momentum (p_final) = -1 * mass * velocity = -1 * 7.5 * 10^-2 kg * 21 m/s = -1.575 kg*m/s
3. Impulse = p_final - p_initial = -1.575 kgm/s - 1.575 kgm/s = -3.15 kg*m/s

However, because the ball hits and rebounds at a 45 degree angle, we need to consider the components of the momentum perpendicular to the wall.

The perpendicular components of the initial and final momenta are 1.575 kgm/s * sin(45) = 1.113 kgm/s and -1.575 kgm/s * sin(45) = -1.113 kgm/s, respectively.

So the change in momentum, or impulse, in the direction perpendicular to the wall is -1.113 kgm/s - 1.113 kgm/s = -2.226 kg*m/s.

Therefore, the magnitude of the impulse given to the ball is 2.226 kg*m/s.