The momentum of block B afterwards is [tex]-8.0 kg m/s[/tex].
Explanation:
We can solve this problem by applying the principle of conservation of momentum.
In fact, the total momentum before and after the collision must be conserved. Therefore, we can write:
[tex]p_A + p_B = p_A' + p_B'[/tex]
where:
[tex]p_A = +15.0 kg m/s[/tex] is the initial momentum of block A
[tex]p_B = -35.0 kg m/s[/tex] is the initial momentum of block B
[tex]p_A' = -12.0 kg m/s[/tex] is the final momentum of block A
[tex]p_B'[/tex] is the final momentum of block B
Solving for [tex]p_B'[/tex], we find:
[tex]p_B' = p_A + p_B - p_A' = +15.0 +(-35.0) - (-12.0)=-8.0 kg m/s[/tex]
So, the momentum of block B afterwards is [tex]-8.0 kg m/s[/tex].
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