Answer:
Explanation:
The momentum of the neutron before and after the decay is the same since there's no external force.
#The neutron is initially at rest, so after the decay:
#After decay, the proton has +ve direction with a velocity while the electron moves in a negative direction with a velocity
Therefore:
Let the energy released during the decay be Q:
Hence,Kp/Ktot is 5.444x10^(-4)
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.
Answer:
The kinetic energy of the baseball is 306.25 joules.
Explanation:
SInce the baseball can be considered a particle, that is, that effects from geometry can be neglected, the kinetic energy (), in joules, is entirely translational, whose formula is:
(1)
Where:
- Mass, in kilograms.
- Speed, in meters per second.
If we know that and
, then the kinetic energy of the baseball thrown by the player is:
The kinetic energy of the baseball is 306.25 joules.
Answer:
Force=35 N
Explanation:
Given data
mass m=10.0 kg
Gravitational field E=3.50 N/kg
To find
Force
Solution
From definition of gravitational field intensity.