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
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]
=> [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]
=> [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]
