To perform the procedure we proceed to define the equations that can help us,
So for the conservation of linear momentum we know that,
[tex]m1_u_1+m_2+u_2= (m_1+m_2)v[/tex]
Clearing for v,
[tex]v=\frac{m_1u_1+m_2u_2}{m_1+m_2}[/tex]
Substituting,
[tex]v= \frac{0.05Kg*v_0+1Kg(0m/s)}{0.05+1}[/tex]
[tex]v=0.0476v_0[/tex]
Where [tex]v_0[/tex] represents the initial velocity of the ball.
For the second part we need to know through kinetic energy the percentage loss, in this way
[tex](KE_i)\% = \frac{KE_{1i}-KE_{1f}}{KE_{1i}} *100[/tex]
Initial
[tex]KE_{1i}=\frac{1}{2}m_1u^2_1[/tex]
Final
[tex]KE_{1f}=\frac{1}{2}(m_1+m_2)v^2[/tex]
Substituting everything in the percentage equation we have
[tex](KE_i)\% = \frac{0.5*(0.05*v_0^2-0.5(0.05+1))*0.0476v_0}{0.5(0.05)*v_0^2}*100[/tex]
[tex](KE_i)\% = \frac{0.05*v_0^2-0.002379*v_0^2}{0.05*v_0^2}*100[/tex]
[tex](KE_i)\% = 95.2\%[/tex]
Which means that this is the percentage of loss of kinetic energy.
The speed of the brick after the collision is 0.048v₀.
The percentage of the mechanical energy lost in the collision is 95.2%.
The given parameters;
Let the speed of the brick after collision = v
Apply the principle of conservation of linear momentum;
[tex]m_1v_0 + m_2 u_2 = v(m_1 + m_2)\\\\0.05v_0 + 0 = v(1 + 0.05)\\\\0.05v_0 = 1.05v\\\\v= \frac{0.05v_0}{1.05} \\\\v = 0.048 v_0[/tex]
Thus, the speed of the brick after the collision is 0.048v₀
The total initial kinetic energy;
[tex]K.E_i = \frac{1}{2} \times 0.05 \times v_0^2 \ + \ \frac{1}{2} \times 0.05 \times 0^2\\\\K.E_i = 0.025v_0^2[/tex]
The total final kinetic energy;
[tex]K.E_f = \frac{1}{2} \times (1.05) \times (0.048v_0)^2\\\\K.E_f = 0.0012 v_0^2[/tex]
The percentage of mechanical energy lost during the collision;
[tex]= \frac{0.025v_0^2 \ - \ 0.0012v_0^2}{0.025v_0^2} \times 100\%\\\\= 95.2 \ \%[/tex]
Learn more here:https://brainly.com/question/24424291
