An atomic nucleus of mass m traveling with speed v collides elastically with a target particle of mass 2m (initially at rest) and is scattered at 90°.

(a) At what angle does the target particle move after the collision?

(b) What are the final speeds of the two particles?

(c) What fraction of the initial kinetic energy is transferred to the target particle?

The momentum and the kinetic energy are conserved during this collision as the collision is elastic.

The initial speed of the atomic nucleus is \(v\) and the initial speed of the target particle iszero as at rest.

After collison, the speed of the atomic nucleus is \({v_1}\) and the speed of the target particle is \({v_2}\) at an angle \(\theta \) with the initial direction of the atomic nucleus.

Part (a)

Using the momentum conservation along the x-axis, you get:

\(\begin{array}{c}mv = 2m{v_2}\cos \theta \\v = 2{v_2}\cos \theta \\{v^2} = 4v_2^2{\cos ^2}\theta \end{array}\) … (i)

Using the momentum conservation along the y-axis, you get:

\(\begin{array}{c}0 = m{v_1} - 2m{v_2}\sin \theta \\{v_1} = 2{v_2}\sin \theta \\v_1^2 = 4v_2^2{\sin ^2}\theta \end{array}\) … (ii)

As the kinetic energy conserved in elastic collison, then:

\(\begin{array}{l}\frac{1}{2}m{v^2} + \left( {\frac{1}{2} \times 2m \times {0^2}} \right) = \frac{1}{2}mv_1^2 + \left( {\frac{1}{2} \times 2m \times v_2^2} \right)\\{v^2} = v_1^2 + 2v_2^2\\{v^2} - v_1^2 = 2v_2^2\end{array}\) … (iii)

Now, adding equations (i) and (ii), you get:

\(\begin{array}{c}{v^2} + v_1^2 = 4v_2^2{\cos ^2}\theta + 4v_2^2{\sin ^2}\theta \\{v^2} + v_1^2 = 4v_2^2\end{array}\) … (iv)

Then, adding equations (iii) and(iv), you get:

\(\begin{array}{c}{v^2} - v_1^2 + {v^2} + v_1^2 = 2v_2^2 + 4v_2^2\\2{v^2} = 6v_2^2\\v_2^2 = \frac{{{v^2}}}{3}\\{v_2} = \frac{v}{{\sqrt 3 }}\end{array}\) … (v)

Now, from equations (i) and (v), you get:

\(\begin{array}{c}{v^2} = 4{\left( {\frac{v}{{\sqrt 3 }}} \right)^2}{\cos ^2}\theta \\{\cos ^2}\theta = \frac{3}{4}\\\cos \theta = \frac{{\sqrt 3 }}{2}\\\theta = {30.0^ \circ }\end{array}\)

Hence, after the collision, the angle of the target particle is \({30.0^ \circ }\)clockwise with the initial direction of the atomic nucleus.

From equations (iv) and (v), you get:

\(\begin{array}{c}{v^2} + v_1^2 = 4{\left( {\frac{v}{{\sqrt 3 }}} \right)^2}\\v_1^2 = \frac{4}{3}{v^2} - {v^2}\\v_1^2 = \frac{{{v^2}}}{3}\\{v_1} = \frac{v}{{\sqrt 3 }}\end{array}\)

Hence, after the collision, both particles are moving at a speed of \(\frac{v}{{\sqrt 3 }}\).

Part (c)

The kinetic energy of the atomic nucleus before collison is \({K_1} = \frac{1}{2}m{v^2}\).

Now, the kinetic energy of the target particle after the collision is:

\(\begin{array}{c}{K_2} = \frac{1}{2} \times 2m \times v_2^2\\ = \frac{1}{2} \times 2m \times {\left( {\frac{v}{{\sqrt 3 }}} \right)^2}\\ = \frac{2}{3} \times \frac{1}{2}m{v^2}\\ = \frac{2}{3}{K_1}\end{array}\)

Hence, \(\frac{2}{3}\) fraction of the initial kinetic energy is transferred to the target particle.