Question 8.11: A ball hits a wall Suppose you throw a ball with mass 0.40 k...
A ball hits a wall
Suppose you throw a ball with mass 0.40 kg against a brick wall (Figure 8.21). It hits the wall moving horizontally to the left at 30 m/s and rebounds horizontally to the right at 20 m/s. Find the impulse of the force exerted on the ball by the wall. If the ball is in contact with the wall for 0.010 s, find the average force on the ball during the impact.
SET UP We take the x axis as horizontal and the positive direction to the right.
SOLVE We need to find the x component of momentum just before and just after impact, find the change in momentum, and then equate this change to the impulse. Then we divide the impulse by the time interval to find the average force on the ball.
The initial x component of momentum of the ball is
p_{i,x}=m\upsilon _{i,x}=(0.40 kg)(-30 m/s)=-12 kg\cdot m/s.
The final x component of momentum is
p_{f,x}=m\upsilon _{f,x}=(0.40 kg)(20 m/s) = 8.0 kg\cdot m/s
The change in the x component of momentum is
\Delta p_x = p_{f,x} – p_{i,x} = m\upsilon _{f,x} – m\upsilon _{i,x}
= 8.0 kg · m/s – (-12 kg · m/s) = 20 kg · m/s.
Note the signs carefully. The initial x component of momentum is negative, and we are subtracting it from the final value to find the change \Delta p_x.) According to Equation 8.17, the preceding result equals the x component of impulse of the force exerted on the ball by the wall, so J_x = 20 kg \cdot m/s = 20 N \cdot s.
If the force acts for 0.010 s, then, from J_x = F_{\mathrm{av,} x} \Delta t, we have
\overrightarrow{\pmb{J}}= \overrightarrow{\pmb{F}}_{\mathrm{av}}(t_f -t_i). (8.17)
F_{\mathrm{av}, x} = \frac{J_x}{\Delta t}=\frac{20 N\cdot s}{0.010 s} = 2000 N.
REFLECT The force exerted by the wall on the ball isn’t constant; its variation with time may be similar to a varying force described by one of the curves in Figure 8.22. The 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 a golf ball, the time of collision is short and the maximum force is large, as in curve A of the figure. If the ball is softer, like a tennis ball, the collision time is longer and the maximum force is less, as in curve B. Either way, the area under the curve represents the same impulse: J_x = F_{\mathrm{av}, x} \Delta t.
Practice Problem: Suppose, rather than throwing a ball, you throw a lump of clay with the same mass and initial velocity as given in this example. When it hits the wall, the lump of clay sticks to the wall. Find the average force exerted on the wall by the lump of clay if it comes to a stop in 0.010 s. Answer: 1200 N.
