A tennis ball of mass m = 0.065 kg and speed v = 35 m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° (see figure). What is the impulse given to the wall?
Magnitude
well, the impulse was given perpendicular to the wall. THe compoent of velocity perpendicular is 35*.707, so change velocity is twice that.
impulse= 2*35*.707 kg m/s
To find the impulse given to the wall, we first need to understand what impulse is. Impulse is the change in momentum of an object and is calculated by multiplying the force acting on the object by the time interval over which the force is applied. Mathematically, impulse (J) can be expressed as:
J = Δp = m (v_f - v_i)
Where Δp is the change in momentum, m is the mass of the object, v_f is the final velocity, and v_i is the initial velocity. In this case, we can assume that the initial velocity is the velocity of the tennis ball before it strikes the wall (v_i = 35 m/s), and the final velocity is the velocity of the tennis ball after it rebounds (v_f = 35 m/s).
Using the given values, we can calculate the impulse given to the wall:
J = m (v_f - v_i)
= (0.065 kg) (35 m/s - 35 m/s)
= 0 Ns
Therefore, the impulse given to the wall is 0 Newton-seconds (Ns).