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A 50 kg archer standing on frictionless ice shoots a 100g arrow at a speed of 100 m/s what uis the recoild pseed of the archer?

1 Answer

3 votes

Answer:

v= 0.2 m/s

Step-by-step explanation:

Given that

m₁ = 50 kg

m₂ = 100 g = 0.1 kg

u =10 0 m/s

If there is no any external force on the system then the total linear momentum of the system will be conserve.

Initial linear momentum = Final momentum

m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁

50 x 0 + 0.1 x 100 = 50 v + 0

0+ 10 = 50 v


v=(10)/(50)\ m/s

v= 0.2 m/s

Therefore the recoil speed will be 0.2 m/s.