AP Physics

Unit 4 - Energy

Intermediate

Mathematical

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The launching mechanism of a toy gun consists of a spring with an unknown spring constant, k k . When the spring is compressed 0.120m 0.120 \, \text{m} vertically, a 35.0g 35.0 \, \text{g} projectile is able to be fired to a maximum height of 25m 25 \, \text{m} above the position of the projectile when the spring is compressed. Assume that the barrel of the gun is frictionless.

  1. (a) Where will you set y=0y = 0? Explain your reasoning. 3 points

  2. (b) Determine the spring constant of the spring. 3 points

  3. (c) Determine the speed that the projectile is traveling the moment it leaves the spring. 3 points

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(a) Where to set y=0 y = 0 :

Step Derivation/Formula Reasoning
1 Setting y=0 y = 0 at the compressed position of the spring By setting y=0 y = 0 at the position where the spring is compressed, we simplify the calculations for the potential energy of the projectile when it is launched. This allows us to treat the initial potential energy as entirely elastic and the final height as purely gravitational potential energy.

(b) Determine the spring constant k k :

Next, let’s calculate the spring constant:

Step Derivation/Formula Reasoning
1 EPE=12kx2EPE = \frac{1}{2}kx^2 Elastic Potential Energy (EPE) stored in the compressed spring. x x is the compression distance.
2 GPE=mghGPE = mgh Gravitational Potential Energy (GPE) at the maximum height h h above the initial position. m m is the mass, g g is gravitational acceleration, and h h is the height.
3 12kx2=mgh\frac{1}{2}kx^2 = mgh Applying conservation of energy, the EPE at the beginning is converted to GPE at the maximum height.
4 k=2mghx2k = \frac{2mgh}{x^2} Solving for the spring constant k k .
5 k=2(0.035kg)(9.8m/s2)(25m)(0.120m)2k = \frac{2 (0.035 \, \text{kg}) (9.8 \, \text{m/s}^2) (25 \, \text{m})}{(0.120 \, \text{m})^2} Substitute the known values: m=0.035kg m = 0.035 \, \text{kg} , g=9.8m/s2 g = 9.8 \, \text{m/s}^2 , h=25m h = 25 \, \text{m} , x=0.120m x = 0.120 \, \text{m} .
6 k=20.0359.8250.0144k = \frac{2 \cdot 0.035 \cdot 9.8 \cdot 25}{0.0144} Calculate the values inside the equation.
7 k=17.150.01441191.67N/mk = \frac{17.15}{0.0144} \approx 1191.67 \, \text{N/m} Divide to find k k . Therefore, the spring constant is approximately 1191.67N/m \boxed{1191.67 \, \text{N/m}} .

(c) Determine the speed that the projectile is traveling the moment it leaves the spring:

Finally, let’s calculate the speed:

Step Derivation/Formula Reasoning
1 EPE=12kx2EPE = \frac{1}{2}kx^2 Using the elastic potential energy stored in the spring.
2 KE=12mv2KE = \frac{1}{2}mv^2 Kinetic Energy (KE) of the projectile at the moment it leaves the spring. m m is the mass and v v is the velocity.
3 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2 Applying conservation of energy, the EPE is converted to KE when the projectile leaves the spring.
4 kx2=mv2kx^2 = mv^2 Eliminate the common factor (1/2) on both sides.
5 v=kx2mv = \sqrt{\frac{kx^2}{m}} Solve for v v , the velocity of the projectile as it leaves the spring.
6 v=1191.67N/m(0.120m)20.035kgv = \sqrt{\frac{1191.67 \, \text{N/m} \cdot (0.120 \, \text{m})^2}{0.035 \, \text{kg}}} Substitute the known values: k=1191.67N/m k = 1191.67 \, \text{N/m} , x=0.120m x = 0.120 \, \text{m} , m=0.035kg m = 0.035 \, \text{kg} .
7 v=1191.670.01440.035v = \sqrt{\frac{1191.67 \cdot 0.0144}{0.035}} Calculate the values inside the equation.
8 v=489.6722.12m/sv = \sqrt{489.67} \approx 22.12 \, \text{m/s} Therefore, the speed of the projectile as it leaves the spring is approximately 22.12m/s \boxed{22.12 \, \text{m/s}} .

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  1. Where the spring is fully compressed
  2. k=17.150.01441191.67N/mk = \frac{17.15}{0.0144} \approx 1191.67 \, \text{N/m}
  3. v=489.6722.12m/sv = \sqrt{489.67} \approx 22.12 \, \text{m/s}