Answer:
Velocity of the helium nuleus = 1.44x10⁴m/s
Velocity of the proton = 2.16x10⁴m/s
Explanation:
From the conservation of linear momentum of the proton collision with the He nucleus:
[tex] P_{1i} + P_{2i} = P_{1f} + P_{2f] [/tex] (1)
where [tex]P_{1i}[/tex]: is the proton linear momentum initial, [tex]P_{2i}[/tex]: is the helium nucleus linear momentum initial, [tex]P_{1f}[/tex]: is the proton linear momentum final, [tex]P_{2f}[/tex]: is the helium nucleus linear momentum final
From (1):
[tex] m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex] (2)
where m₁ and m₂: are the proton and helium mass, respectively, [tex]v_{1i}[/tex] and [tex]v_{2i}[/tex]: are the proton and helium nucleus velocities, respectively, before the collision, and [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex]: are the proton and helium nucleus velocities, respectively, after the collision
By conservation of energy, we have:
[tex] K_{1i} + K_{2i} = K_{1f} + K_{2f} [/tex] (3)
where [tex]K_{1i}[/tex] and [tex]K_{2i}[/tex]: are the kinetic energy for the proton and helium, respectively, before the colission, and [tex]K_{1f}[/tex] and [tex]K_{2f}[/tex]: are the kinetic energy for the proton and helium, respectively, after the colission
From (3):
[tex] \frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] (4)
Now we have two equations: (2) ad (4), and two incognits: [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex].
Solving equation (2) for [tex]v_{1f}[/tex], we have:
[tex] v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} [/tex] (5)
From getting (5) into (4) we can obtain the [tex]v_{2f}[/tex]:
[tex] v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0 [/tex]
[tex] v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0 [/tex]
From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:
[tex] v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} [/tex] (6)
Now, by introducing (6) into (5) we get the proton velocity after the collision:
[tex] v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4} [/tex]
[tex] v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s} [/tex]
The negative sign means that the proton is moving in the opposite direction after the collision.
I hope it helps you!
