Respuesta :
Answer:
Explanation:
mass of proton = m
initial velocity, u1 = 1.80 x 10^7 m/s
final velocity, v1 = - 1.5 x 10^7 m/s
(a)
Let m' is the mass of unknown nucleus.
By use of conservation of energy
[tex]m'=\frac{u_{1}-v_{1}}{u_{1}+v_{1}}m[/tex]
[tex]m'=\frac{1.8\times 10^{7}-(-1.5\times 10^{7})}{1.8\times 10^{7}+(-1.5\times 10^{7})}m[/tex]
m' = 11 m
Thus, the mass of unknown nucleus is 11 times the mass of proton.
(b)
By use of conservation of momentum
m x u1 + m' x 0 = m x v1 + m' x v'
where, v' is the velocity of unknown nucleus after collision
m x 1.8 x 10^7 = - m x 1.5 x 10^7 + 11 m x v'
3.3 m x 10^7 = 11 m x v'
v' = 3 x 10^6 m/s
Thus, the velocity of unknown nucleus after collision is 3 x 10^6 m/s
A) The mass of one nucleus of the unknown element in terms of the proton mass m is;
m₂ = 11m
B) The speed of the unknown nucleus immediately after such a given collision is;
v₂ = 3 × 10^(6) m/s
The missing parts of the question are;
(a) Find the mass of one nucleus of the unknown element. Express your answer in terms of the proton mass m.
(b) What is the speed of the unknown nucleus immediately after such a collision?
We are given;
Mass of proton = m
Initial speed of protons; u₁ = 1.8 × 10^(7) m/s
Final speed of protons; v₁ = -1.5 × 10^(7) m/s (It's negative because it rebounds)
A) Since the collision is elastic and we are told that the initial speed of the target nucleus is negligible, by law of conservation of energy, we can say that;
m₂/m = (u₁ - v₁)/(u₁ + v₁)
Where;
m₂ is the mass of one nucleus of the unknown element.
Thus, Plugging in the relevant values, we have;
m₂/m = [(1.8 + 1.5)10^(7)]/[(1.8 - 1.5)10^(7)]
m₂/m = 3.3/0.3
m₂/m = 11
Thus;
m₂ = 11m
B) We want to find the speed of the unknown nucleus immediately after such a collision.
From conservation of linear momentum, we will have the formula;
mu₁ + m₂u₂ = mv₁ + m₂v₂
Now, we are told that initial speed of the target nucleus is negligible. Thus; u₂ = 0 m/s
Plugging in other relevant values gives;
m(1.8 × 10^(7)) = m(-1.5 × 10^(7)) + 11mv₂
m will cancel out all through to give;
1.8 × 10^(7) = (-1.5 × 10^(7)) + 11v₂
11v₂ = (1.8 + 1.5) × 10^(7)
11v₂ = 3.3 × 10^(7)
v₂ = (3.3/11) × 10^(7)
v₂ = 3 × 10^(6) m/s
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