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Anni [7]
2 years ago
6

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 speed. What is the mass of the second ball
Physics
1 answer:
emmasim [6.3K]2 years ago
6 0

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)

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A system initially has an internal energy U of 504 J. It undergoes a process during which it releases 111 J of heat energy to th
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You apply a potential difference of 5.70 v between the ends of a wire that is 2.90 m in length and 0.654 mm in radius. the resul
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V=IR
where V is the potential difference applied on the wire, I the current and R the resistance. For the resistor in the problem we have:
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2) Now that we have the value of the resistance, we can find the resistivity of the wire \rho by using the following relationship:
\rho =  \frac{RA}{L}
Where A is the cross-sectional area of the wire and L its length.
We already have its length L=2.90 m, while we need to calculate the area A starting from the radius:
A=\pi r^2 = \pi (0.654\cdot 10^{-3}m)^2=1.34 \cdot 10^{-6}m^2

And now we can find the resistivity:
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