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:
Answer:
v= 0.2 m/s
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= 0.2 m/s
Therefore the recoil speed will be 0.2 m/s.
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Answer:
The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.
Explanation:
The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.
To solve this problem, start with setting up the net force equations for both block A and B:
where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:
Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:
The force of friction acting on block B is approximately 26.7N.
This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.