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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Answer:
240 meter
Explanation:
d= s*d=60m/s * 4sec=240 meter
Explanation:
The given data is as follows.
k = 130 N/m,
= 17 cm = 0.17 m (as 1 m = 100 cm)
mass (m) = 2.8 kg
When the spring is compressed then energy stored in it is as follows.
Energy = 
Now, spring energy gets converted into kinetic energy when the box is launched.
So,
= 
= 

= 1.34
v = 1.15 m/sec
Now,
Frictional force = 
= 
= 4.116 N
Also, Kinetic energy = work done by friction
1.8515 =
d = 0.449 m
Thus, we can conclude that the box slides 0.449 m across the rough surface before stopping.