The momentum of the 2.22 kg afterwards is 0.91 kg m/s to the right
Explanation:
We can solve this problem by applying the law of conservation of momentum.
In fact, the total momentum before and after the collision must be conserved. 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 = 3.00 kg[/tex] is the mass of the first block
[tex]u_1 = 2.09 m/s[/tex] is the initial velocity of the first block (we take the right as positive direction)
[tex]v_1 = -1.11 m/s[/tex] is the final velocity of the first block (to the left)
[tex]m_2 = 2.22 kg[/tex] is the mass of the second block
[tex]u_2 = -3.92[/tex] is the initial velocity of the second block
[tex]v_2[/tex] is the final velocity of the second block
Re-arranging the equation and substituting the values, we find: the final velocity of the second block:
[tex]v_2 = \frac{m_1 u_1+m_2 u_2 - m_1 v_1}{m_2}=\frac{(3.00)(2.09)+(2.22)(-3.92)-(3.00)(-1.11)}{2.22}=0.41 m/s[/tex]
(to the right, since it is positive)
And so, the momentum of the 2.22 kg block afterwards is:
[tex]p_2 = m_2 v_2 = (2.22)(0.41)=0.91 kg m/s[/tex] (to the right)
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