The momentum of the 155 kg car afterwards is 469.7 kg m/s to the right
Explanation:
We can solve the problem by using the law of conservation of momentum: the total momentum of the system is conserved before and after the collision, so we can write:
[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2[/tex]
where:
[tex]m_1 = 282 kg[/tex] is the mass of the bumper car
[tex]u_1 = 3.50 m/s[/tex] is the initial velocity of the bumper car (we take the right as positive direction)
[tex]v_1 = 0.800 m/s[/tex] is the final velocity of the bumper car
[tex]m_2 = 155 kg[/tex] is the mass of the second bumper car
[tex]u_2 = -1.88 m/s[/tex] is the initial velocity of the second car (moving to the left)
[tex]v_2[/tex] is the final velocity of the second car
Solving for [tex]v_2[/tex],
[tex]v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(282)(3.50)+(155)(-1.88)-(282)(0.800)}{155}=3.03 m/s[/tex]
where the positive sign means the direction is to the right.
And now we can find the momentum of the 155 kg afterwards, which is
[tex]p_2 = m_2 v_2 = (155)(3.03)=469.7 kg m/s[/tex] (to the right)
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