Answer:
a) the mass of the second car = 28000 kg
b) the amount of kinetic energy that was lost in the collision = [tex]246.9*10^3 \ J[/tex]
Explanation:
Given that:
mass of the train car [tex]m_1[/tex] = 6300 kg
speed of the train car [tex]v_1[/tex] = 12.0 m/s
mass of the second moving car [tex]m_2[/tex] = ???
speed of the second moving car [tex]v_2[/tex] = 2.2 m/s
After strike;
they both move with a speed [tex]v_f[/tex] = 4.00 m/s
a)
Using the conservation of momentum :
[tex]m_1v_1+m_2v_2 = (m_1 + m_2)v_f[/tex]
[tex](6300*12)+m_2(2.2) = (6300 + m_2)4[/tex]
[tex](75600)+m_2(2.2) = 25200 + 4m_2[/tex]
[tex]75600 - 25200 = 4m_2 -2.2m_2[/tex]
[tex]50400 = 1.8m_2[/tex]
[tex]m_2 = \frac{50400}{1.8}[/tex]
[tex]m_2 = 28000 \ kg[/tex]
b)
To determine the amount of kinetic energy that was lost in the collision;
we will need to find the difference between the kinetic energy before the collision and after the collision;
i.e
[tex]K.E _{lost} = K.E_{i} - K.E _{f}[/tex]
[tex]K.E_{i} = \frac{1}{2} m_1v_1^2 + \frac{1}{2}m_2v_2^2[/tex]
[tex]K.E_{i} = \frac{1}{2} (6300)(12)^2 + \frac{1}{2}(28000)(2.2)^2[/tex]
[tex]K.E_{i} = 521360 \ J[/tex]
[tex]K.E_{f} = \frac{1}{2}(m_1 + m_2 ) v_f ^2[/tex]
[tex]K.E_{f} = \frac{1}{2}(6300 + 28000 ) (4)^2[/tex]
[tex]K.E_{f} =274400 \ J[/tex]
Now; the kinetic energy that was lost in the collision is calculated as follows:
[tex]K.E _{lost} = K.E_{i} - K.E _{f}[/tex]
[tex]K.E_{lost} = (521360-274400) \ J[/tex]
[tex]K.E_{lost} =246960 \ J[/tex]
[tex]K.E_{lost} =246.9 * 10 ^3 \ J[/tex]
