Question

a proton of mass 1 u travelling with a speed of 3.6 x 10 ^4 m/s has an elastic head on collision with a helium nucleus initially at rest. Assuming the collision takes place in nearly empty space (that is no external forces are involved so momentum will be conserved), what are the velocities of the proton and helium after the collision?

Answer 1

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:

$P_{1i} + P_{2i} = P_{1f} + P_{2f]$ (1)

where $P_{1i}$: is the proton linear momentum initial, $P_{2i}$: is the helium nucleus linear momentum initial, $P_{1f}$: is the proton linear momentum final, $P_{2f}$: is the helium nucleus linear momentum final

From (1):

$m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f}$ (2)

where m₁ and m₂: are the proton and helium mass, respectively, $v_{1i}$ and $v_{2i}$: are the proton and helium nucleus velocities, respectively, before the collision, and $v_{1f}$ and $v_{2f}$: are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

$K_{1i} + K_{2i} = K_{1f} + K_{2f}$ (3)

where $K_{1i}$ and  $K_{2i}$: are the kinetic energy for the proton and helium, respectively, before the colission, and $K_{1f}$ and  $K_{2f}$: are the kinetic energy for the proton and helium, respectively, after the colission

From (3):

$\frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2}$ (4)  

Now we have two equations: (2) ad (4), and two incognits: $v_{1f}$ and $v_{2f}$.

Solving equation (2) for $v_{1f}$, we have:

$v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f}$ (5)

From getting (5) into (4) we can obtain the $v_{2f}$:

$v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0$

$v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0$

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

$v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s}$ (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

$v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4}$

$v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s}$

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!