Question

A spring-loaded dart gun is compressed a distance d, then fired straight up into the air. The dart reaches a maximum height of 24 m. For a second trial, the spring is compressed a distance of 0.50d. What is the maximum height of the dart in the second trial? Assume that the effect of friction and other nonconservative forces is negligible.

Answer 1

Answer: 12 m

Since the compression became smaller hence the height reached will become smaller proportionally

Answer 2

Answer: 6m

Explanation:

1) The assumption that the efferct of friction and other noncoservative forces is negligible, leads you to state that the work done by the gun is completely transformed into gravitational potential energy of the dart.

2) The work exerted by the gun is equal to the change of elastic potential energy of the gun, which is:

W = - k Δx²

The first time, that was W = - kd²

3) The change in gravitational potential energy of the dart, the first time, was:

ΔPE = mg ΔH = mg(24) = 24mg

Therefore: - kd² = 24mg ↔ equation 1

4) The second time, the work exerted by the gun was:

W = - k Δx²= - k(0.5d)²

5) The change in gravitational potential energy of the dart the second was:

ΔPE = mg ΔH = mgh

Therefore: - k(0.5)d² mgh ↔ equation 2

6) Divide equation 2 by equation 1:

k(0.5d)² / [kd²] = mgh /[24mg] → now cancel h, d², m, and g.

⇒ (0.5)² = h/24

⇒ h = 24(0.5)² = 6

Therefore, the answer is 6 m.