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Question* An a-particle moving at a speed of 1.0×106 ms−11.0 \times 10^{6} \mathrm{~ms}^{-1} collides with a stationary proton. After willitin The proton travels with a speed of 1.88×106ir1.88 \times 10^{6} \mathrm{ir} in the divectuon in which the GG-particle. war travelling. Calculate the boed of the xx-partels offer collim. [Marie of α\alpha-partido =6.64×10−239=6.64 \times 10^{-23} 9 Nas of prite =166×10−64 kg=166 \times 10^{-64} \mathrm{~kg}

Studdy Solution

STEP 1

1. The collision is perfectly elastic.
2. The initial velocity of the proton is zero, as it is stationary.
3. The conservation of momentum applies to this collision.

STEP 2

1. Recall the conservation of momentum formula.
2. Substitute the given values into the momentum equation.
3. Solve for the speed of the alpha particle after the collision.

STEP 3

Recall the conservation of momentum formula:
For a two-body system, the total momentum before the collision equals the total momentum after the collision. Mathematically, this is expressed as:
m1v1i+m2v2i=m1v1f+m2v2f m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}
where: - m1 m_1 and m2 m_2 are the masses of the alpha particle and proton, respectively. - v1i v_{1i} and v2i v_{2i} are the initial velocities of the alpha particle and proton, respectively. - v1f v_{1f} and v2f v_{2f} are the final velocities of the alpha particle and proton, respectively.

STEP 4

Substitute the given values into the momentum equation:
Given: - m1=6.64×10−27 kg m_1 = 6.64 \times 10^{-27} \, \text{kg} (mass of alpha particle) - m2=1.66×10−27 kg m_2 = 1.66 \times 10^{-27} \, \text{kg} (mass of proton) - v1i=1.0×106 m/s v_{1i} = 1.0 \times 10^{6} \, \text{m/s} (initial velocity of alpha particle) - v2i=0 m/s v_{2i} = 0 \, \text{m/s} (initial velocity of proton) - v2f=1.88×106 m/s v_{2f} = 1.88 \times 10^{6} \, \text{m/s} (final velocity of proton)
The equation becomes:
(6.64×10−27 kg)×(1.0×106 m/s)+(1.66×10−27 kg)×0=(6.64×10−27 kg)×v1f+(1.66×10−27 kg)×(1.88×106 m/s) (6.64 \times 10^{-27} \, \text{kg}) \times (1.0 \times 10^{6} \, \text{m/s}) + (1.66 \times 10^{-27} \, \text{kg}) \times 0 = (6.64 \times 10^{-27} \, \text{kg}) \times v_{1f} + (1.66 \times 10^{-27} \, \text{kg}) \times (1.88 \times 10^{6} \, \text{m/s})

STEP 5

Solve for the speed of the alpha particle after the collision (v1f v_{1f} ):
First, simplify the equation:
6.64×10−21=6.64×10−27×v1f+3.1208×10−21 6.64 \times 10^{-21} = 6.64 \times 10^{-27} \times v_{1f} + 3.1208 \times 10^{-21}
Rearrange to solve for v1f v_{1f} :
6.64×10−21−3.1208×10−21=6.64×10−27×v1f 6.64 \times 10^{-21} - 3.1208 \times 10^{-21} = 6.64 \times 10^{-27} \times v_{1f}
3.5192×10−21=6.64×10−27×v1f 3.5192 \times 10^{-21} = 6.64 \times 10^{-27} \times v_{1f}
v1f=3.5192×10−216.64×10−27 v_{1f} = \frac{3.5192 \times 10^{-21}}{6.64 \times 10^{-27}}
v1f≈5.30×105 m/s v_{1f} \approx 5.30 \times 10^{5} \, \text{m/s}
The speed of the alpha particle after the collision is approximately:
5.30×105 m/s \boxed{5.30 \times 10^{5} \, \text{m/s}}

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