A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
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Answer:
(a) the final velocity of the cart is 0.857 m/s
(b) the impulse experienced by the package is 21.43 kg.m/s
(c) the fraction of the initial energy lost is 0.71
Explanation:
Given;
mass of the package, m₁ = 10 kg
mass of the cart, m₂ = 25 kg
initial velocity of the package, u₁ = 3 m/s
initial velocity of the cart, u₂ = 0
let the final velocity of the cart = v
(a) Apply the principle of conservation of linear momentum to determine common final velocity for ineleastic collision;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
10 x 3 + 25 x 0 = v(10 + 25)
30 = 35v
v = 30 / 35
v = 0.857 m/s
(b) the impulse experienced by the package;
The impulse = change in momentum of the package
J = ΔP = m₁v - m₁u₁
J = m₁(v - u₁)
J = 10(0.857 - 3)
J = -21.43 kg.m/s
the magnitude of the impulse experienced by the package = 21.43 kg.m/s
(c)
the initial kinetic energy of the package is calculated as;
the final kinetic energy of the package;
the fraction of the initial energy lost;