The surface of the inclined plane shown is frictionless. If F = 30 N, what is the
1 answer:
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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:
20J
Explanation:
In a collision, whether elastic or inelastic, momentum is always conserved. Therefore, using the principle of conservation of momentum we can first get the final velocity of the two bodies after collision. This is given by;
m₁u₁ + m₂u₂ = (m₁ + m₂)v ---------------(i)
Where;
m₁ and m₂ are the masses of first and second objects respectively
u₁ and u₂ are the initial velocities of the first and second objects respectively
v is the final velocity of the two objects after collision;
From the question;
m₁ = 2.0kg
m₂ = 8.0kg
u₁ = 5.0m/s
u₂ = 0 (since the object is initially at rest)
<em>Substitute these values into equation (i) as follows;</em>
(2.0 x 5.0) + (8.0 x 0) = (2.0 + 8.0)v
(10.0) + (0) = (10.0)v
10.0 = 10.0v
v = 1m/s
The two bodies stick together and move off with a velocity of 1m/s after collision.
The kinetic energy(KE₁) of the objects before collision is given by
KE₁ = m₁u₁² +
m₂u₂² ---------------(ii)
Substitute the appropriate values into equation (ii)
KE₁ = ( x 2.0 x 5.0²) + (
x 8.0 x 0²)
KE₁ = 25.0J
Also, the kinetic energy(KE₂) of the objects after collision is given by
KE₂ = (m₁ + m₂)v² ---------------(iii)
Substitute the appropriate values into equation (iii)
KE₂ = ( 2.0 + 8.0) x 1²
KE₂ = 5J
The kinetic energy lost (K) by the system is therefore the difference between the kinetic energy before collision and kinetic energy after collision
K = KE₂ - KE₁
K = 5 - 25
K = -20J
The negative sign shows that energy was lost. The kinetic energy lost by the system is 20J