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.
<h3>What is the law of conservation of momentum?</h3>
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;

Hence, the final velocity of the block A will be 2.5 m/sec.
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The potential energy of the block is given by:
V = m*g*h
m mass
g = 9.81m/s²
h height
The potential energy of a spring is given by:
V = 0.5 * k * x²
k spring constant
x compression of the spring
If the block starts from rest it has potential energy, but no kinetic energy. As it slides down the incline potential energy is converted into kinetic energy. When the block hits the spring the kinetic energy is converted into spring's potential energy. If the spring is fully compressed and the block is at rest again, the block has transferred all its energy into the spring. No energy is lost. So we can write:
m * g * h = 0.5 * k * x²
m = 0.5 kg
g = 9.81 m/s²
h = 2.5m * sin 37° = 1,5 m
x = 0,6 m
Solve for k.
k = 2 * m * g * h / x² = 40.8 N/m
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