a 3.00 kg block moving 2.09 m/s right hits a 2.22 kg block moving 3.92 m/s left. afterward, the 3.00 kg…

Explanation:

Step1: Calculate initial momentum of 3.00 kg block

The formula for momentum is $p = mv$. For the 3.00 kg block, $m_1 = 3.00$ kg and $v_1=2.09$ m/s, so $p_1=m_1v_1 = 3.00\times2.09=6.27$ kg·m/s.

Step2: Calculate initial momentum of 2.22 kg block

For the 2.22 kg block, $m_2 = 2.22$ kg and $v_2=- 3.92$ m/s, so $p_2=m_2v_2=2.22\times(-3.92)= - 8.6024$ kg·m/s.

Step3: Calculate final momentum of 3.00 kg block

The 3.00 kg block moves at $v_{1f}=-1.11$ m/s after the collision. So $p_{1f}=m_1v_{1f}=3.00\times(-1.11)=-3.33$ kg·m/s.

Step4: Use conservation of momentum

According to the law of conservation of momentum $p_{1}+p_{2}=p_{1f}+p_{2f}$. Rearranging for $p_{2f}$ gives $p_{2f}=p_{1}+p_{2}-p_{1f}$. Substitute the values: $p_{2f}=6.27+( - 8.6024)-(-3.33)=6.27 - 8.6024 + 3.33 = 0.9976\approx1.00$ kg·m/s.

Answer:

$1.00$ kg·m/s