Question 8.3: The Spring-Loaded Popgun The launching mechanism of a popgun......
The Spring-Loaded Popgun
The launching mechanism of a popgun consists of a trigger-released spring (Active Fig. 8.6a on page 208). The spring is compressed to a position y_{Ⓐ}, and the trigger is fired. The projectile of mass m rises to a position y_{Ⓒ} above the position at which it leaves the spring, indicated in Active Figure 8.6b as position y_{Ⓑ} = 0. Consider a firing of the gun for which m = 35.0 \mathrm{~g}, y_{Ⓐ}=-0.120 \mathrm{~m}, and y_{Ⓒ}=20.0 \mathrm{~m}.
(A) Neglecting all resistive forces, determine the spring constant.
(B) Find the speed of the projectile as it moves through the equilibrium position Ⓑ of the spring as shown in Active Figure 8.6b.
(A) Conceptualize Imagine the process illustrated in parts (a) and (b) of Active Figure 8.6. The projectile starts from rest, speeds up as the spring pushes upward on it, leaves the spring, and then slows down as the gravitational force pulls downward on it.
Categorize We identify the system as the projectile, the spring, and the Earth. We ignore air resistance on the projectile and friction in the gun, so we model the system as isolated with no nonconservative forces acting.
Analyze Because the projectile starts from rest, its initial kinetic energy is zero. We choose the zero configuration for the gravitational potential energy of the system to be when the projectile leaves the spring. For this configuration, the elastic potential energy is also zero.
After the gun is fired, the projectile rises to a maximum height y_{Ⓒ}. The final kinetic energy of the projectile is zero.
From the isolated system model, write a conservation of mechanical energy equation for the system between points Ⓐ and Ⓒ:
K_{Ⓒ}+U_{g Ⓒ}+U_{s Ⓒ}=K_{Ⓐ}+U_{g Ⓐ}+U_{s Ⓐ}
Substitute for each energy:
0+m g y_{Ⓒ}+0=0+m g y_{Ⓐ}+\frac{1}{2} k x^{2}
Solve for k :
k=\frac{2 m g\left(y_{Ⓒ}-y_{Ⓐ}\right)}{x^{2}}
Substitute numerical values:
k=\frac{2(0.0350 \mathrm{~kg})\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)[20.0 \mathrm{~m}-(-0.120 \mathrm{~m})]}{(0.120 \mathrm{~m})^{2}}=958 \mathrm{~N} / \mathrm{m}
(B) Analyze The energy of the system as the projectile moves through the equilibrium position of the spring includes only the kinetic energy of the projectile \frac{1}{2} m v_{Ⓑ}{ }^{2}. Both types of potential energy are equal to zero for this configuration of the system.
Write a conservation of mechanical energy equation for the system between points Ⓐ and Ⓑ:
K_{Ⓑ}+U_{g Ⓑ}+U_{s Ⓑ}=K_{Ⓐ}+U_gⒶ+U_{s Ⓐ}
Substitute for each energy:
\frac{1}{2} m v_{Ⓑ}{ }^{2}+0+0=0+m g y_{Ⓐ}+\frac{1}{2} k x^{2}
Solve for v_{Ⓑ} :
v_{Ⓑ}=\sqrt{\frac{k x^{2}}{m}+2 g y_{Ⓐ}}
Substitute numerical values:
v_{Ⓑ}=\sqrt{\frac{(958 \mathrm{~N} / \mathrm{m})(0.120 \mathrm{~m})^{2}}{(0.0350 \mathrm{~kg})}+2\left(9.80 \mathrm{~m} / \mathrm{s}^{2}\right)(-0.120 \mathrm{~m})}=19.8 \mathrm{~m} / \mathrm{s}
Finalize This example is the first one we have seen in which we must include two different types of potential energy. Notice in part (A) that we never needed to consider anything about the speed of the ball between points Ⓐ and Ⓒ, which is part of the power of the energy approach: changes in kinetic and potential energy depend only on the initial and final values, not on what happens between the configurations corresponding to these values.