Mickey, a daredevil mouse of mass M, is attempting to become the world\’s first \”mouse cannonball.\” He is loaded into a spring-powered gun pointing up at some angle, and is shot into the air. The gun\’s spring has force constant k and is initially compressed a distance X from its relaxed position. If Mickey has a constant horizontal speed v while he is flying through the air, how high above his initial location in the gun does Mickey soar? Find an expression for Mickey\’s maximum height h_m in terms of M, v, X, k, and g.
The Correct Answer and Explanation is :
To determine the maximum height hmh_m that Mickey reaches above his initial position in the spring-powered gun, we can analyze the problem using principles of energy conservation and projectile motion.
Energy Conservation:
Initially, the spring is compressed by a distance XX, storing potential energy given by:
PEspring=12kX2PE_{\text{spring}} = \frac{1}{2} k X^2
where kk is the spring constant.
Upon release, this potential energy is converted into Mickey’s kinetic energy. Assuming no energy losses, the total kinetic energy imparted to Mickey is:
KE=12Mv2KE = \frac{1}{2} M v^2
where MM is Mickey’s mass and vv is his constant horizontal speed during flight.
By equating the spring’s potential energy to Mickey’s kinetic energy, we get:
12kX2=12Mv2\frac{1}{2} k X^2 = \frac{1}{2} M v^2
Solving for vv:
v=kMXv = \sqrt{\frac{k}{M}} X
Projectile Motion:
In projectile motion, the horizontal and vertical components of motion are independent. Given that Mickey’s horizontal speed vv remains constant, we can determine the vertical component of his initial velocity vyv_y using trigonometry. If θ\theta is the launch angle:
vy=vsinθv_y = v \sin \theta
The maximum height hmh_m above the launch point is reached when the vertical component of Mickey’s velocity becomes zero. The formula for the maximum height in projectile motion is:
hm=vy22gh_m = \frac{v_y^2}{2g}
Substituting vyv_y:
hm=(vsinθ)22gh_m = \frac{(v \sin \theta)^2}{2g}
Since v=kMXv = \sqrt{\frac{k}{M}} X, we have:
hm=(kMXsinθ)22gh_m = \frac{\left( \sqrt{\frac{k}{M}} X \sin \theta \right)^2}{2g}
Simplifying:
hm=kX2sin2θ2Mgh_m = \frac{k X^2 \sin^2 \theta}{2 M g}
Conclusion:
The maximum height hmh_m that Mickey reaches above his initial position in the gun is:
hm=kX2sin2θ2Mgh_m = \frac{k X^2 \sin^2 \theta}{2 M g}
This expression shows that the maximum height depends on the spring constant kk, the compression distance XX, Mickey’s mass MM, the gravitational acceleration gg, and the square of the sine of the launch angle θ\theta. Notably, the horizontal speed vv is related to these parameters but does not appear explicitly in the final expression for hmh_m.