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? 1) Mechanical energy is conserved because no dissipative forces perform work on the ball. 2) The forces of gravity and the spring have potential energies associated with them. 3) No conservative forces act in this problem after the ball is released from the spring gun.

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).

Part C Find the maximum height hmax of the ball.

Part D Which of the following actions, if done independently, would increase the maximum height reached by the ball? 1) reducing the spring constant k 2) increasing the spring constant k 3) decreasing the distance the spring is compressed 4) increasing the distance the spring is compressed 5) decreasing the mass of the ball 6) increasing the mass of the ball 7) tilting the spring gun so that it is at an angle θ<90 degrees from the horizontal

Answer 1

Given that :

mass of the ball, m = 1.5 kg

spring constant, k = 667 N/m

spring compression distance, x = 25 cm = 0.25 m

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 : The true statements are given as -

1. Mechanical energy is conserved because no dissipative forces perform work on the ball.

2. The force of gravity & spring have potential energies associated with them.

Part-B : The muzzle velocity of the ball which is given as :

using conservation of energy, we have

K.E = P.E_spring

(1/2) k x² = (1/2) m v₀²

k x² = m v₀² { eq.1 }

inserting the values in eq.1,

(667 N/m) (0.25 m)² = (1.5 kg) v₀²

(41.6 Nm) = (1.5 kg) v₀²

v₀ = 27.7 m²/s²

v₀ = 5.26 m/s

using equation of motion 3, we have

v² = v₀² + 2 a x { eq.2 }

inserting the values in eq.2,

v² = (5.26 m/s)² + 2 (-9.8 m/s²) (0.25 m)

v² = (27.6 m²/s²) - (4.9 m²/s²)

v = 22.7 m²/s²

v = 4.76 m/s

Part-C : The maximum height of the ball which will be given as -

using an equation,   v² = 2 g h_max                                                          { eq.3 }

inserting the values in eq.3,

(4.76 m/s)² = 2 (9.8 m/s²) h_max

(22.6 m²/s²) = (19.6 m/s²) h_max

h_max = (22.6 m²/s²) / (19.6 m/s²)

h_max = 1.15 m

Part-D : The following actions would increase the maximum height reached by the ball which is given as -

2. increasing the spring constant k

4. increasing the distance the spring is compressed

5. decreasing the mass of the ball