Question

10. A 5.0 g pellet is placed in the barrel of a toy gun and is propelled by a spring of force
constant 50. N/m. The spring is compressed 20. cm and then released. Calculate the
maximum velocity of the pellet when shot horizontally.

Answer 1

Answer:

v = 20m/s

Explanation:

m = 5g = 0.005kg

k = 50 N/m

e = 20cm = 0.20 m

Ek = Eep

1/2 mv² = 1/2 ke²

mv² = ke²

v² = ke²/m

v = √(ke²/m)

v = √400

v = 20m/s

Answer 2

The maximum velocity of the pellet is 282.8 m/s.

Note: The kinetic energy of the pellet is equal to the spring's potential energy.

To calculate the maximum velocity of the pellet, we use the formula below.

Formula:

• mv²/2 = ke²/2.
• mv² = ke²................ Equation 1

Where:

• m = mass of the pellet
• v = velocity of the pellet'
• k = spring constant of the pellet
• e = extension.

From the question:

Given:

• m = 5 g = 0.005 kg
• k = 50 N/m
• e = 20 cm = 0.2 m

Substitute these values into equation 1

• 0.005v² = 50(0.2²)

Solve for v

• v² = 50(0.2²)/(0.005²)
• v² = 80000
• v = √(80000)
• v = √(80000)
• v = 282.8 m/s.

Hence the maximum velocity of the pellet is 282.8 m/s.

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