An alpha particle collides with an oxygen nucleus, initially at rest. The alpha particle is scattered at an angle of 60.9° above its initial direction of motion, and the oxygen nucleus recoils at an angle of 51.9° on the opposite side of that initial direction. The final speed of the nucleus is 1.32 105 m/s. In atomic mass units, the mass of an alpha particle is 4.0 u. The mass of an oxygen nucleus is 16 u.
(a) Find the final speed of the alpha particle.
m/s
(b) Find the initial speed of the alpha particle.
m/s
I tried this and got
118882 for Va'
and 390097 for Va
but it was wrong?
To solve this problem, we can use conservation of momentum and conservation of kinetic energy.
Let's begin with part (a) to find the final speed of the alpha particle.
Given:
Mass of alpha particle (m1) = 4.0 u
Mass of oxygen nucleus (m2) = 16 u
Angle of scatter for alpha particle (θ1) = 60.9°
Angle of recoil for oxygen nucleus (θ2) = 51.9°
Final speed of oxygen nucleus (v2) = 1.32 x 10^5 m/s
To find the final speed of the alpha particle (v1), we can use the conservation of momentum.
Conservation of momentum:
m1v1i + m2v2i = m1v1f + m2v2f
The initial speeds of both particles are zero (initially at rest), so the equation becomes:
m1v1f = m2v2f
Now let's substitute the given values into the equation:
4.0 u * v1f = 16 u * 1.32 x 10^5 m/s
To solve for v1f, we need to convert the atomic mass unit (u) to kg and cancel out the units:
4.0 u * 1.66 x 10^-27 kg/u * v1f = 16 u * 1.32 x 10^5 m/s
Now, we can solve for v1f:
v1f = (16 u * 1.32 x 10^5 m/s) / (4.0 u * 1.66 x 10^-27 kg/u)
Calculating the result:
v1f ≈ 2.43 x 10^7 m/s
Therefore, the final speed of the alpha particle (v1) is approximately 2.43 x 10^7 m/s.
Moving on to part (b), let's find the initial speed of the alpha particle (v1i).
To find the initial speed of the alpha particle, we will now use the conservation of kinetic energy.
Conservation of kinetic energy:
(1/2) m1 v1i^2 + (1/2) m2 v2i^2 = (1/2) m1 v1f^2 + (1/2) m2 v2f^2
Since the particles are initially at rest, the equation becomes:
(1/2) m1 v1i^2 = (1/2) m1 v1f^2 + (1/2) m2 v2f^2
Now let's substitute the given values into the equation:
(1/2) * 4.0 u * v1i^2 = (1/2) * 4.0 u * (2.43 x 10^7 m/s)^2 + (1/2) * 16 u * (1.32 x 10^5 m/s)^2
To solve for v1i, we'll convert the atomic mass unit (u) to kg and cancel out the units:
(1/2) * 4.0 u * 1.66 x 10^-27 kg/u * v1i^2 = (1/2) * 4.0 u * (2.43 x 10^7 m/s)^2 + (1/2) * 16 u * (1.32 x 10^5 m/s)^2
Now, we can solve for v1i:
v1i^2 = [(4.0 u * (2.43 x 10^7 m/s)^2) + (16 u * (1.32 x 10^5 m/s)^2)] / (4.0 u * 1.66 x 10^-27 kg/u)
Calculating the result:
v1i ≈ 5.82 x 10^6 m/s
Therefore, the initial speed of the alpha particle (v1i) is approximately 5.82 x 10^6 m/s.
To solve this problem, we need to apply the principle of conservation of momentum and conservation of kinetic energy.
(a) Find the final speed of the alpha particle:
1. Start by finding the magnitude of the final velocity of the oxygen nucleus:
Given:
Mass of alpha particle (m1) = 4.0 u
Mass of oxygen nucleus (m2) = 16 u
Final velocity of the oxygen nucleus (v2) = 1.32 x 10^5 m/s
Using the conservation of momentum formula:
(m1 * v1) + (m2 * v2) = m1 * v1' + m2 * v2'
v1' = [(m1 * v1) + (m2 * v2) - (m2 * v2')] / m1
Since the initial velocity of the oxygen nucleus is zero, the equation simplifies to:
v1' = (m2 * v2') / m1
v1' = (16 u * 1.32 x 10^5 m/s) / 4.0 u
v1' = 5.28 x 10^5 m/s
2. To find the final speed of the alpha particle (v1'), we can use the conservation of kinetic energy.
The kinetic energy of the alpha particle before the collision (K1) is given by:
K1 = (1/2) * m1 * v1^2
The kinetic energy of the oxygen nucleus after the collision (K2') is given by:
K2' = (1/2) * m2 * v2'^2
Using the conservation of kinetic energy formula:
K1 = K2'
(1/2) * m1 * v1^2 = (1/2) * m2 * v2'^2
Solving for v1:
v1 = √[(m2 * v2'^2 * m1) / m2]
v1 = √[(1.32 x 10^5 m/s * 16 u * 4.0 u) / 16 u]
v1 = √(4.224 x 10^6 m^2/s^2)
v1 = 2.06 x 10^3 m/s
So, the final speed of the alpha particle is approximately 2.06 x 10^3 m/s.
(b) Find the initial speed of the alpha particle:
To find the initial speed of the alpha particle (v1), we can use the conservation of momentum:
The total momentum of the system before the collision is zero since both the alpha particle and oxygen nucleus are initially at rest.
(m1 * v1) + (m2 * v2) = 0
Solving for v1:
v1 = -[(m2 * v2)] / m1
v1 = -[(16 u * 1.32 x 10^5 m/s)] / 4.0 u
v1 = -5.28 x 10^5 m/s
Since speed cannot be negative, the initial speed of the alpha particle is approximately 5.28 x 10^5 m/s.
The final momentum components of the two particles perpendicular to the initial direction must cancel. Therefore
1.32*10^5*sin51.9 = Va'*sin60.9
Solve for Va', the final speed of the alpha particle.
Get the initial speed of the alpha particle, Va, by applying conservation of momentum to the intial direction of motion.
Va*4 = 16*1.32*10^5*cos51.0 + 4*Va'*cos60.9
You know Va' from part a. Solve for Va