A spring-loaded toy gun is used to shoot a ball of mass m =1.50kg straight up in

Question

A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in (Figure 1) . The spring has spring constant k=667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height hmax (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.

Part A

Which of the following statements are true?

Check all that apply.

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Correct

Part B

Find vm the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y=0).

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Part C

Find the maximum height hmax of the ball.

Express your answer numerically, in meters.

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A spring-loaded toy gun is used to shoot a ball of mass m=1.50kg straight up in the air, as shown in (Figure 1) . The spring has spring constant k=667N/m. If the spring is compressed a distance of 25.0 centimeters from its equilibrium position y=0 and then released, the ball reaches a maximum height hmax (measured from the equilibrium position of the spring). There is no air resistance, and the ball never touches the inside of the gun. Assume that all movement occurs in a straight line up and down along the y axis.

Part A

Which of the following statements are true?

Check all that apply.

Mechanical energy is conserved because no dissipative forces perform work on the ball. The forces of gravity and the spring have potential energies associated with them. No conservative forces act in this problem after the ball is released from the spring gun.

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Correct

Part B

Find vm the muzzle velocity of the ball (i.e., the velocity of the ball at the spring's equilibrium position y=0).

vm =   m/s  

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Part C

Find the maximum height hmax of the ball.

Express your answer numerically, in meters.

hmax =   m  

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max. y 30

Explanation / Answer

Question is not present

1
a)true
b)true
c)false, gravity is a conservative force

2.energy before = energy after
Ki + Ui + Si = Kf + Uf + Sf

where K is kinetic, U is potential, and S is spring energy. Some of these can be eliminated for the first part. Their equations are

K = 1/2*m*v^2
U = m*g*h
S = 1/2*k*x^2

Initial velocity is zero, so Ki = 0. Final position is at y = 0, so Uf = 0. The spring is unstretched in the final position, giving Sf = 0. This leaves:

Ui + Si = Kf
m*g*h + 1/2*k*x^2 = 1/2*m*v^2

You know everything except for v, solve. Remember that h is negative, your ball is below the y = 0 line initially.

Now for the second part, apply the same equation. This time, the initial position is the instant the spring is in its unstretched position, and final position is at maximum height.

Lets eliminate some things again. The initial position is at y = 0, so Ui = 0. There is no spring involved in this part, so Si and Sf = 0. At the max height, v = 0, so Kf = 0. This leaves:

Ki = Uf
1/2*m*v^2 = m*g*h

where v is the muzzle velocity calculated above.

first find energy of the spring with
E=1/2*k*x^2 = 20.84 J
assuming its 100% efficient and all work is transferred as kinetic energy, we can use the equation
KE=1/2*m*v^2
v=sqrt(2*Ke/m) = 5.27 m/s

3.
we use
2*a*x=v^2-v0^2
x=(v^2-v0^2)/(2*a) = v0^2/(2*g) = 0.289 m

a) reducing the spring constant k - no
b) increasing the spring constant k - yes
c) decreasing the distance the spring is compressed - no
d) increasing the distance the spring is compressed - yes
e) decreasing the mass of the ball - yes
f) increasing the mass of the ball - no
g) tilting the spring gun so that it is at an angle theta < 90 degrees from the horizontal - no

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