The final velocity of the block A will be 2.5 m/sec. The principal of the momentum conversation is used in the given problem.
According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.
In a given concern, mass m₁ is M, mass m₂ is 3M. Initial speed for the mass m₁ and m₂ will be u₁=5 and u₂=0 m/s respectively,
According to the law of conservation of momentum
Momentum before collision =Momentum after collision
m₁u₁+m₂u₂=(m₁+m₂)v
M×5+3M×0=[M+3M]v
The final velocity is found as;
V=51.25 m/s
The velocity of block A is found as;
[tex]\rm V_f=\frac{(m_1-m_2)u_1}{m_1+m_2} +\frac{2m_2u_2}{m_1+m_2} \\\\ V_f=\frac{(M-3M)5+2\times3M\times0}{m_1+m_2} \\\\ V_f=2.5 m/s[/tex]
Hence, the final velocity of the block A will be 2.5 m/sec.
To learn more about the law of conservation of momentum, refer;
https://brainly.com/question/1113396
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