Question 8.2: A BALL HITS A WALL You throw a ball with a mass of 0.40 kg ...

IDENTIFY and SET UP:

We’re given enough information to determine the initial and final values of the ball’s momentum, so we can use the impulse–momentum theorem to find the impulse. We’ll then use the definition of impulse to determine the average force. Figure 8.6 shows our sketch. We need only a single axis because the motion is purely horizontal. We’ll take the positive x-direction to be to the right. In part (a) our target variable is the x-component of impulse, J_x, which we’ll find by using Eqs. (8.9). In part (b), our target variable is the average x-component of force \left(F_{\mathrm{av}}\right)_{x}; once we know J_x, we can also find this force by using Eqs. (8.9):

(J_{x}=\int_{t_{1}}^{t_{2}} \Sigma F_{x} d t=\left(F_{\mathrm{av}}\right)_{x}\left(t_{2}-t_{1}\right)=p_{2 x}-p_{1 x}=m v_{2 x}-m v_{1 x}     ,      J_{y}=\int_{t_{1}}^{t_{2}} \Sigma F_{y} d t=\left(F_{\mathrm{av}}\right)_{y}\left(t_{2}-t_{1}\right)=p_{2 y}-p_{1 y}=m v_{2 y}-m v_{1 y})

EXECUTE:

(a) With our choice of x-axis, the initial and final x-components of momentum of the ball are

From the x-equation in Eqs. (8.9), the x-component of impulse equals the change in the x-momentum:

J_{x}=p_{2 x}-p_{1 x}=8.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}-(-12 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})=20 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}=20 \mathrm{~N} \cdot \mathrm{s}

(b) The collision time is t_{2}-t_{1}=\Delta t=0.010 \mathrm{~s}. From the x-equation in Eqs. (8.9), J_{x}=\left(F_{\mathrm{av}}\right)_{x}\left(t_{2}-t_{1}\right)=\left(F_{\mathrm{av}}\right)_{x} \Delta t, So

EVALUATE: The x-component of impulse J_x is positive—that is, to the right in Fig. 8.6. The impulse represents the “kick” that the wall imparts to the ball, and this “kick” is certainly to the right.

CAUTION: momentum is a vector Because momentum is a vector, we had to include the negative sign in writing p_{1x} = -12 kg * m/s. Had we omitted it, we would have calculated the impulse to be 8.0 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}-(12 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s})=-4 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}. This would say that the wall had somehow given the ball a kick to the left! Remember the direction of momentum in your calculations.

The force that the wall exerts on the ball must have such a large magnitude (2000 N, equal to the weight of a 200-kg object) to change the ball’s momentum in such a short time. Other forces that act on the ball during the collision are comparatively weak; for instance, the gravitational force is only 3.9 N. Thus, during the short time that the collision lasts, we can ignore all other forces on the ball. Figure 8.7 shows the impact of a tennis ball and racket.
Note that the 2000-N value we calculated is the average horizontal force that the wall exerts on the ball during the impact. It corresponds to the horizontal line \left(F_{\mathrm{av}}\right)_{x} in Fig. 8.3a. The horizontal force is zero before impact, rises to a maximum, and then decreases to zero when the ball loses contact with the wall. If the ball is relatively rigid, like a baseball or golf ball, the collision lasts a short time and the maximum force is large, as in the blue curve in Fig. 8.3b. If the ball is softer, like a tennis ball, the collision time is longer and the maximum force is less, as in the orange curve in Fig. 8.3b.