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7. A ball of mass m makes a head-on elastic collision with a second ball (at rest) and rebounds with a speed equal to 0.450 its original spe
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Answer:
mass of the second ball is 0.379m
Explanation:
Given;
mass of first ball = m
let initial velocity of first ball = u₁
let final velocity of first ball = v₁ = 0.45u₁
let the mass of the second ball = m₂
initial velocity of the second ball, u₂ = 0
let the final velocity of the second ball = v₂
Apply the principle of conservation of linear momentum;
mu₁ + m₂u₂ = mv₁ + m₂v₂
mu₁ + 0 = 0.45u₁m + m₂v₂
mu₁ = 0.45u₁m + m₂v₂ ——– equation (i)
Velocity for elastic collision in one dimension;
u₁ + v₁ = u₂ + v₂
u₁ + 0.45u₁ = 0 + v₂
1.45u₁ = v₂ (final velocity of the second ball)
Substitute in v₂ into equation (i)
mu₁ = 0.45u₁m + m₂(1.45u₁)
mu₁ = 0.45u₁m + 1.45m₂u₁
mu₁ – 0.45u₁m = 1.45m₂u₁
0.55mu₁ = 1.45m₂u₁
divide both sides by u₁
0.55m = 1.45m₂
m₂ = 0.55m / 1.45
m₂ = 0.379m
Therefore, mass of the second ball is 0.379m (where m is mass of the first ball)