Answer
The magnitude of the impulse applied to the golf ball by the floor is $3.66N.m/s$
Work Step by Step
In this exercise, we need to look at the horizontal and vertical components of the momentum separately.
- On the horizontal side, the momentum $\vec{p}_x$ does not change during the impact, so the floor applies no impulse to the golf ball on this side, according to the impulse-momentum theorem.
- On the vertical, take the upward movement to be the positive direction.
We have $\vec{p_0}=m\vec{v_0}\cos30=0.047\times(-45)\cos30=-1.83kg.m/s$
$\vec{p_f}=m\vec{v_f}\cos30=0.047\times(+45)\cos30=+1.83kg.m/s$
According to the impulse-momentum theorem, $$\vec{J}=\vec{p}_f-\vec{p}_0=+3.66N.m/s$$ $$J=3.66N.m/s$$
