Answer:
v = 1.4 m / s
Explanation:
To solve this problem we use the law of conservation of momentum, for this we define a system formed by the two blocks in such a way that the force during the collision have been internal
initial instant. Just before the crash
p₀ = M v₁ - m v₂
final instant. Right after the crash
p_f = (M + m) v
the moment is preserved
p₀ = p_f
M v₁ - m v₂ = (M + m) v
v = [tex]\frac{1}{M+m}[/tex] (M v₁ - mv₂)
let's calculate
v = [tex]\frac{1}{4+1}[/tex] (4 2 - 1 1)
v = 1.4 m / s
in the same direction as the largest block (M)
The approximate speed of block X after the collision is 1.4 m/s.
Given information:
Block X and block Y travel toward each other along a horizontal surface with block X traveling in the positive direction.
Block X has a mass of [tex]m_1=[/tex] 4 kg and a speed of [tex]u_1=[/tex] 2 m/s.
Block Y has a mass of [tex]m_2=[/tex] 1 kg and a speed of [tex]u_2=[/tex] -1 m/s.
The collision between them is perfectly inelastic. So, the final velocity of both the objects will be the same.
The momentum of the system will be conserved. Let v be the final velocity of both the blocks.
The final velocity v can be calculated as,
[tex]m_1u_1+m_2u_2=(m_1+m_2)v\\4\times 2-1\times1=(4+1)v\\5v=7\\v=1.4\rm\;m/s[/tex]
Therefore, the approximate speed of block X after the collision is 1.4 m/s.
For more details about collision, refer to the link:
https://brainly.com/question/12941951
