Answer :
Answer:
-3.6 m/s
Explanation:
Here we need to use the conservation of momentum law.
So, the momentum of the man initially is zero and so of the object. The momentum of the object after throw is 5 kg multiplied by 50 m/s which is 250 kg m/s.
So the equation is as follows:
0 + 0 = (250 + 70 x v) kg m/s
Which leads to v = (-250/70) m/s ≈ - 3.6 m/s
I am not 100 % sure though! Hope this helps! :)
The recoil velocity of the man obtained is –3.6 m/s (Option A)
Data obtained from the question
- Mass of man (m₁) = 70 Kg
- Mass of object (m₂) = 5 Kg
- Velocity of object (v₂) = 50 m/s
- Velocity of man (v₁) =?
How to determine the velocity of the man
The recoil velocity of the man can be obtained as illustrated below
m₁v₁ = –m₂v₂
70 × v₁ = –(5 × 50)
70 × v₁ = –250
Divide both side by 70
v₁ = –250 / 70
v₁ = –3.6 m/s
Therefore, the recoil velocity of the man is –3.6 m/s
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