Question 9.38: In the overhead view of Fig. 9-54, a 300 g ball with a speed......
In the overhead view of Fig. 9-54, a 300 g ball with a speed ν of 6.0 m/s strikes a wall at an angle θ of 30° and then rebounds with the same speed and angle. It is in contact with the wall for 10 ms. In unitvector notation, what are (a) the impulse on the ball from the wall and (b) the average force on the wall from the ball?
From Fig. 9-54, +y corresponds to the direction of the rebound (directly away from the wall) and +x toward the right. Using unit-vector notation, the ball’s initial and final velocities are
{\vec{\nu}}_{i}=\nu\cos\theta{\hat{{i}}}-\nu\sin\theta{\hat{{j}}}=5.2\,{\hat{{i}}}-3.0{\hat{{j}}}
\vec{\nu}_{f}=\nu\cos\theta\,\hat{{i}}+\nu\sin\theta\hat{{j}}=5.2\,\hat{{i}}+3.0\hat{{j}}
respectively (with SI units understood).
(a) With m = 0.30 kg, the impulse-momentum theorem (Eq. 9-31) yields
\Delta\vec{p}=\vec{J} (linear momentum-impulse theorem). (9-31)
{\vec{J}}=m{\vec{\nu}}_{f}-m{\vec{\nu}}_{i}=2(0.30 kg)(3.0 m/s \hat{j})=(1.8 N·s)\hat{j}.
(b) Using Eq. 9-35, the force on the ball by the wall is {\vec{J}}/{\Delta}t=(1.8/0.010){\hat{{j}}}=(180\,\mathrm{N}){\hat{{j}}}. By Newton’s third law, the force on the wall by the ball is (-180\ \mathrm{N})\hat j − (that is, its magnitude is 180 N and its direction is directly into the wall, or “down” in the view provided by Fig. 9-54).
J=F_{\mathrm{avg}}\,\Delta t. (9-35)