The momentum of block B after the collision is -50 kg m/s.
Explanation:
We can solve this problem by using the principle of conservation of momentum. In fact, the total momentum of the system before and after the collision must be conserved, so we can write:
[tex]p_A + p_B = p'_A + p'_B[/tex]
where:
[tex]p_A = -100 kg m/s[/tex] is the momentum of block A before the collision
[tex]p_B = -150 kg m/s[/tex] is the momentum of block B before the collision
[tex]p'_A = -200 kg m/s[/tex] is the momentum of block A after the collision
[tex]p'_B[/tex] is the momentum of block B after the collision
Solving for [tex]p'_B[/tex], we find:
[tex]p'_B = p_A + p_B - p'_A = -100 +(-150) - (-200)=-50 kg m/s[/tex]
So, the momentum of block B after the collision is -50 kg m/s.
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