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A 10-M-Long Glider With A Mass Of 680 Kg (Including The Passengers) Is Gliding Horizontally Through The Air

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A 10-m-long glider with a mass of 680 kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 60 kg skydiver drops out by releasing his grip on the glider. What is the glider’s velocity just after the skydiver lets go?

 

Known:

π‘šπ‘† = 60 π‘˜π‘” (mass of skydiver)

π‘šπΊ+𝑆 = 680 π‘˜π‘” (mass of glider plus the mass of skydiver)

π‘šπΊ = π‘šπΊ+𝑆 − π‘šπ‘† = 680 π‘˜π‘” – 60 π‘˜π‘” = 620 π‘˜π‘” (mass of glider)

𝑣𝐺𝑖 = 30 m/s

π‘šπ‘† = 60π‘˜π‘”

Note: turns out as the skydiver releases, the skydiver’s final velocity will be the same as glider’s initial velocity thus,

𝑣𝑆𝑓 = 𝑣𝐺i

Find:

𝑣𝐺𝑓 =?

 

SOLUTION:

Using the Law of Conservation of Momentum equation (𝑷𝒇 = π‘·π’Š),

(π‘šπΊ)𝑣𝐺𝑓 + (π‘šπ‘†)𝑣𝑆𝑓 = (π‘šπΊ+𝑆)𝑣𝐺𝑖

Substitute 𝑣𝐺𝑖 for 𝑣𝑆f given that 𝑣𝑆f = 𝑣𝐺𝑖, then solve for the gliders final velocity

(π‘šπΊ)𝑣𝐺𝑓 + (π‘šπ‘†)𝑣𝐺𝑖= (π‘šπΊ+𝑆)𝑣𝐺i

(mG)vGf = (mG+S)vGi – (mS)vGi

vGf = (mG+S)vGi – (mS)vGi / (mG)

vGf = (680)30 – (60)30 / (620)

vGf = 30 m/s

The glider’s velocity just after the skydiver lets go is 30 m/s.

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