Answer:
The mass of the second ball is [tex]m_2 = 2.175m \ kg[/tex]
Explanation:
From the question we are told that
The mass of the first ball is m
Let assume
The original speed of the first ball is [tex]u_1[/tex]
The rebound speed is [tex]v_1 = 0.370u_1 \ m/s[/tex]
According to the law of momentum conservation
[tex]m_1 u_1 + m_2 u_2 = -m_1 v_1 + m_2 v_2[/tex]
The second ball is at rest so
[tex]u_2 = 0[/tex]
And the negative sign shows that it is a rebound velocity
[tex]mu_1 + = -m v_1 + m_2 v_2[/tex]
=> [tex]mu_1 + = -0.370 mu_1 + m_2 v_2[/tex]
Now we are told that the collision is elastic, this means that the velocity of approach will be equal to the velocity of separation
So
[tex]u_1 = v_1 + v_2[/tex]
=> [tex]u_1 = 0.370u_1 + v_2[/tex]
=> [tex]v_2 = 0.63 \ u_1[/tex]
substituting this into equation above
[tex]mu_1 + = -0.370 mu_1 + m_2 (0.63 u_1)[/tex]
[tex]1.370 mu_1 = m_2 (0.63 u_1)[/tex]
=> [tex]m_2 = 2.175m \ kg[/tex]
The mass of the second ball in terms of the first ball, m, is 2.175 m.
The given parameters;
Apply the principle of conservation of linear momentum to determine the mass of the second ball as shown below;
[tex]m_1 u_1 + m_2 u_2 = m_1v_1 + m_2 v_2\\\\mu_1 + m_2(0) = m(-0.37u_1) + m_2v_2\\\\mu_1 = -0.37u_1m + m_2v_2\\\\mu_1 + 0.37mu_1 = m_2v_2\\\\1.37 mu_1 = m_2 v_2 \ --(1)[/tex]
Apply one-dimensional velocity equation for elastic collision;
[tex]u_1 + v_1 = u_2 + v_2\\\\u_1 + (-0.37u_1) = 0 + v_2\\\\u_1 -0.37u_1 = v_2\\\\v_2 = 0.63 u_1[/tex]
substitute the value of v₂ into the above equation;
[tex]1.37mu_1 = m_2(0.63u_1)\\\\1.37mu_1 = 0.63m_2u_1\\\\m_2 = \frac{1.37mu_1}{0.63u_1} \\\\m_2 = 2.175 \ m[/tex]
Thus, the mass of the second ball in terms of the first ball, m, is 2.175 m.
Learn more here:https://brainly.com/question/3227235
