2.10-kg box is moving to the right with speed 8.50 m>s on a horizontal, frictionless surface. At t = 0 a horizontal force is
applied to the box. The force is directed to the left and has magnitude F1t2 = 16.00 N>s22t2. (a) What distance does the box move from its position at t = 0 before its speed is reduced to zero? (b) If the force continues to be applied, what is the speed of the box at t = 3.00 s?
2 answers:
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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.