Answer:
v₂ = - 1.29 m / s
, K₁ / Em = 0.996
, K₂ / Em = 0.004
Explanation:
This is an exercise for the moment, where the system is formed by the shotgun, the man arm and the bullet, in this case the forces are internal therefore the moment is conserved
Initial. Before shooting
p₀ = 0
Final. After shooting
= m v₁ + M v₂
Where index 1 is for the bullet and index 2 for the shotgun, arm and man set
The mass of the bullet is m = 0.04 kg
The mass of the set M = 3.0 + 12.5 = 15.5 kg
p₀ =
0 = m v₁ + M v₂
v₂ = - m / M v₁
v₂ = - 0.04 / 15.5 400
v₂ = - 1.29 m / s
The mechanical energy is equal to the kinetic energy, let's calculate the energy for each of the two elements
K₁ = ½ m v₁²
K₁ = 1/2 0.04 400²
K₁ = 3200 J
K₂ = ½ M v₂²
K₂ = ½ 15.5 1.29²
K₂ = 12.90 J
The total energy of the system is the sum of the energy of each component
Em = K₁ + K₂
Em = 3200 + 12.90
Em = 3212.9 J
The fraction of transmitted energy is
To the bullet
K₁ / Em = 3200 / 3212.9
K₁ / Em = 0.996
To the system
K₂ / Em = 12.9 / 3212.9
K₂ / Em = 0.004
The velocity of the ball is 12.5 m/s
Explanation:
The velocity of the ball is given by the ratio between the distance covered by the ball and the time taken:
First, we calculate the distance covered. We know that the radius of the circle is
r = 0.450 m
And the length of the circumference is
The ball makes 25.0 revolutions, so a total distance of
In a time of
t = 9.37 s
So, its velocity is
Learn more about velocity here:
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