Answer:
a
The mass is  [tex]m_2 =21.75*10^{-27} \ kg[/tex]
b
The velocity is  [tex]v_2 = 3.0*10^{6} m/s[/tex]
Explanation:
From the question we are told that
   The speed of the protons is  [tex]u_1 = 2.10*10^{7} m/s[/tex]
   The mass of the protons is  [tex]m[/tex]
   The speed of the rebounding protons are [tex]v_1 = -1.80 * 10^{7} \ m/s[/tex]
The negative sign shows that it is moving in the opposite direction
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Now according to the law of energy conservation mass of one nucleus of the unknown element. is mathematically represented as
    [tex]m_2 = [\frac{u_1 -v_1}{u_1 + v_1} ] m_1[/tex]
Where [tex]m_1[/tex] is the mass of a single proton
     So substituting values
    [tex]m_2 = \frac{2.10 *10^{7} - (-1.80 *10^{7})} {(2.10 *10^7) + (-1.80 *10^{7})} m_1[/tex]
    [tex]m_2 =13 m_1[/tex]
The mass of on proton is  [tex]m_1 = 1.673 * 10^{-27} \ kg[/tex]
So   [tex]m_2 =13 ( 1.673 * 10^{-27} )[/tex]
    [tex]m_2 =21.75*10^{-27} \ kg[/tex]
Now according to the law of linear momentum conservation the speed of the unknown nucleus immediately after such a collision is mathematically evaluated as
   [tex]m_1 u_1 + m_2u_2 = m_1 v_1 + m_2v_2[/tex]
Now  [tex]u_2[/tex] because before collision the the nucleus was at rest
So
    [tex]m_1 u_1 = m_1 v_1 + m_2v_2[/tex]
=> Â Â [tex]v_2 = \frac{m_1(u_1 -v_1)}{m_2}[/tex]
Recall that [tex]m_2 =13 m_1[/tex]
So
    [tex]v_2 = \frac{m_1(u_1 -v_1)}{13m_1}[/tex]
=> Â Â Â Â [tex]v_2 = \frac{(u_1 -v_1)}{13}[/tex]
substituting values
       [tex]v_2 = \frac{( 2.10*10^{7} -(-1.80 *10^{7}))}{13}[/tex]
       [tex]v_2 = 3.0*10^{6} m/s[/tex]
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