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)
Answer:
Explanation:
Given the following data;
Mass = 35 kg
Velocity = 3 m/s
To find the kinetic energy of the child;
K.E = ½mv²
Answer:
Twice
Explanation:
From the formula for velocity in a circle
V= 2πr/T
Where V is velocity
r is raduis
T is period
We see that as r increases V increases so if r is doubled V becomes doubled
