A 10-M-Long Glider With A Mass Of 680 Kg (Including The Passengers) Is Gliding Horizontally Through The Air
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:
𝑚𝑆 = 60 𝑘𝑔 (mass of skydiver)
𝑚𝐺+𝑆 = 680 𝑘𝑔 (mass of glider plus the mass of skydiver)
𝑚𝐺 = 𝑚𝐺+𝑆 − 𝑚𝑆 = 680 𝑘𝑔 – 60 𝑘𝑔 = 620 𝑘𝑔 (mass of glider)
𝑣𝐺𝑖 = 30 m/s
𝑚𝑆 = 60𝑘𝑔
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 = 𝑷𝒊),
(𝑚𝐺)𝑣𝐺𝑓 + (𝑚𝑆)𝑣𝑆𝑓 = (𝑚𝐺+𝑆)𝑣𝐺𝑖
Substitute 𝑣𝐺𝑖
for 𝑣𝑆f given that 𝑣𝑆f = 𝑣𝐺𝑖, then solve for the gliders final velocity
(𝑚𝐺)𝑣𝐺𝑓 + (𝑚𝑆)𝑣𝐺𝑖= (𝑚𝐺+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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