you push a book a distance of 5 meters with a force of 10 newtons for 2 seconds how much work did you do on the book
1 answer:
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Answer:
a) 0.9995c
b) 5641MeV
c) 91670 MeV
Explanation:
(a) The speed of approach is given by the formula:
(b) the kinetic energy is given by:
by replacing c=3*10^8m/s, m_0=1.67*10^{-27}kg we obtain:
(c) in the rest frame of the other proton we have:
by replacing we get
hope this helps!!
Answer:
1) elastic shock, the velocity of the center of mass does not change
2) inelastic shock, he velocity of the mass center change
Explanation:
The position of the center of mass of your system is defined by
=
in this case we have two bodies
x_{cm} = (x₁m₁ + x₂ m₂)
the velocity of the center of mass is
x_{cm} = dx_{cm} / dt =
x_{cm} =
where M is the total mass of the system.
Therefore to answer this question we have to find the velocity of the body after the collision.
Let's use momentum conservation, where the system is formed by the two bodies, so that the forces have been internal during the collision.
Let's solve each case separately.
2) inelastic shock
initial instant. Before the crash
p₀ = m₁ v₀ + 0
final instant. After the collision with the cars together
p_f = (m₁ + m₂) v
p₀ = p_f
m₁ v₀ = (m₁ + m₂) v
v = v₀
let's find the velocity of the center of mass
M = m₁ + m₂
initial.
=
(m₁ vo)
final
=
( v) = v
v_{cm f} =
Let's find the ratio of the velocities of the center of mass
vcmf / vcmo =
therefore the velocity of the mass center change
1) elastic shock
initial instant.
p₀ = m₁ v₀
final moment
p_f = m₁ v_{1f} + m₂ v_{2f}
p₀ = p_f
m₁ v₀ = m₁ v_{1f} + m₂ v_{2f}
m₁ (v₀ - v_{2f}) = m₂ v_{2f}
in this case the kinetic energy is conserved
K₀ = K_f
½ m₁ v₀² = ½ m₁ v_{1f}² + ½ m₂ v_{2f}²
m₁ (v₀² - v_{1f}²) = m₂ v_{2f}²
m₁ (v₀ + v_{1f}) (v₀ - v_{1f}) = m₂ v_{2f}
we write our system of equations
m₁ (v₀ - v_{1f}) = m₂ v_{2f} (1)
m₁ (v₀ - v_{1f}) (v₀ + v_{1f}) = m₂ v_{2f}²
we solve the system
v₀ + v_{1f} = v_{2f}
we substitute and look for the final speeds
v_{1f} =
v_{2f} =
now let's find the velocity of the center of mass
initial
=
m₁ v₀
final
=
(m₁ v_{1f} + m₂ v_{2f} )
v_{cm f} = [
+
] v₀
v_{cm f} = ( m₁² - m₁m₂ +2 m₁m₂) v₂
v_{cm f} = (m₁² + m₁ m₂) v₀
let's look for the relationship
v_{cm f} / v_{cm o} = M
v_{cm f} / v_{cm o} = 1
therefore the velocity of the center of mass does not change
we see in either case the velocity of the center of mass does not change.