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Chapter 7: Problem 66

In a nuclear reactor, a neutron moving at speed \(v_{\mathrm{i}}\) in the positive \(x\)-direction strikes a deuteron, which is at rest. The neutron is deflected by \(90.0^{\circ}\) and moves off with speed $v_{\mathrm{i}} / \sqrt{3}\( in the positive \)y\(-direction. Find the \)x\(- and \)y$-components of the deuteron's velocity after the collision. (The mass of the deuteron is twice the mass of the neutron.)

Short Answer

Expert verified
Answer: The x-component of the deuteron's velocity after the collision is \(v_{Dx}=\frac{1}{2}v_{\text{i}}\), and the y-component of deuteron's velocity is \(v_{Dy}=-\frac{1}{2}\frac{v_{\text{i}}}{\sqrt{3}}\).

Step by step solution

01

List the known values

Given values: - Initial neutron velocity, \(v_{\text{i}}\), in the positive \(x\)-direction - Final neutron velocity, \(\frac{v_{\text{i}}}{\sqrt{3}}\), deflected by \(90^\circ\) and moves off in the positive \(y\)-direction - Mass of deuteron (\(m_D\)) is twice the mass of the neutron (\(m_N\))
02

Express total momentum before and after collision

Before the collision, only the neutron is moving in the positive \(x\)-direction, so the total momentum is given by \(m_Nv_{\text{i}}\). After the collision, both the neutron and deuteron have velocities, which we can express the components as \(v_{Nx}, v_{Ny}, v_{Dx}, v_{Dy}\). The total momentum in the x-direction before and after the collision is: $$m_Nv_{\text{i}}=m_Nv_{Nx}+m_Dv_{Dx}$$ The total momentum in the y-direction before and after the collision is: $$0=m_Nv_{Ny}+m_Dv_{Dy}$$
03

Write the known final velocities of neutron

After the collision, we know the final velocity of the neutron in the \(y\)-direction is \(\frac{v_{\text{i}}}{\sqrt{3}}\). Since the neutron is deflected by \(90^\circ\), this means the final velocity of the neutron in x-direction would be zero. We can plug these values into the equations from Step 2. In the x-direction: $$m_Nv_{\text{i}}=m_N(0)+m_Dv_{Dx}$$ In the y-direction: $$0=m_N\frac{v_{\text{i}}}{\sqrt{3}}+m_Dv_{Dy}$$
04

Solve for the deuteron's velocity components

We can now solve the equations from Step 3 for the deuteron's velocity components (\(v_{Dx}\) and \(v_{Dy}\)). In the x-direction: $$v_{Dx}=\frac{m_N}{m_D}v_{\text{i}}$$ Since the mass of the deuteron is twice the mass of the neutron: $$v_{Dx}=\frac{1}{2}v_{\text{i}}$$ In the y-direction: $$v_{Dy}=-\frac{m_N}{m_D}\frac{v_{\text{i}}}{\sqrt{3}}$$ Again, since the mass of the deuteron is twice the mass of the neutron: $$v_{Dy}=-\frac{1}{2}\frac{v_{\text{i}}}{\sqrt{3}}$$ So, the x- and y-components of the deuteron's velocity after the collision are \(v_{Dx}=\frac{1}{2}v_{\text{i}}\) and \(v_{Dy}=-\frac{1}{2}\frac{v_{\text{i}}}{\sqrt{3}}\) respectively.

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