Two current carrying wires are parallel to one another and separated by 2cm. If the distance between them is increased to 4cm th
1 answer:
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Answer:
v₁=0.2 ms
Explanation:
Given that
m₁ = 0.1 kg
u₁ = 0.2 m/s
m₂ = 0.15 kg
u₂ = 0 m/s
Lets take the final speed of the 0.1 kg ball = v₁
Lets take the final speed of the 0.15 kg ball = v₂
Given that collision is elastic that is why e = 1
u₁ - u₂ = v₁ - v₂
0.2 - 0 = v₁ - v₂
v₁ - v₂ = 0.2 ------------1
If there is no any external force on the system then the total linear momentum of the system will be conserve.
m₁u₁ + m₂ u₂ =m₂ v₂ +m₁v₁
0.2 x 0.1 + 0 = 0.15 v₂ + 0.1 v₁
0.02 = 0.1 v₁ + 0.15 v₂
2 = 10 v₁ + 15 v₂ ----------------2
By using equation 1 and 2
We can say that
v₁=0.2 ms
v₂ =0 m/s