A 57 g tennis ball strikes hits a wall at 12 m/s at an angle of 25° relative to the normal line perpendicular to the wall, and rebounds with the same speed. Find the magnitude of the impulse delivered by the wall.

To solve this problem, we need to use the concept of impulse, which is the change in momentum of an object when a force is applied over a period of time. The formula for impulse (J) is:

J = Δp = m * Δv

where:

• Δp is the change in momentum,
• m is the mass of the object, and
• Δv is the change in velocity.

The change in velocity is the final velocity minus the initial velocity. In this case, the tennis ball hits the wall and rebounds with the same speed but in the opposite direction. Therefore, the change in velocity is twice the initial velocity.

However, since the ball hits and rebounds at an angle, we need to consider the components of the velocity perpendicular to the wall. The initial velocity component perpendicular to the wall is v_i = 12 m/s * cos(25°), and the final velocity component perpendicular to the wall is v_f = -12 m/s * cos(25°).

Therefore, the change in velocity perpendicular to the wall is Δv = v_f - v_i = -12 m/s * cos(25°) - 12 m/s * cos(25°) = -24 m/s * cos(25°).

Substituting the values into the impulse formula gives:

J = m * Δv = 57 g * -24 m/s * cos(25°) = -1368 g*m/s * cos(25°).

However, we need to convert the mass from grams to kilograms (since 1 kg = 1000 g), which gives:

J = -1.368 kgm/s * cos(25°) = -1.24 kgm/s.

Therefore, the magnitude of the impulse delivered by the wall is 1.24 kg*m/s.