"Two blocks of masses m1 and m2 are connected by a spring of spring constant k (figure 9-E15). The block of mass m2 is given a sharp impulse so that it acquires a velocity v0 towards right. Find
(a) the velocity of the center of mass,
(b) the maximum elongation that the spring will suffer."
Consider the situation of the previous problem. Suppose each of the blocks is pulled by a constant force F instead of any impulse. Find the maximum elongation that the spring will suffer and the distances moved by the two blocks in the process
Answers
Thanks for asking the question!
ANSWER::
As both the blocks are pulled by same force , they suddenly move with some acceleration and instantaneously stop at same position where elongation of spring is maximum.
Let extension of block m₁ and m₂ be x₁ and x₂
Total work done = Fx₁ + Fx₂ [Equation 1]
Increase Potential Energy of spring = (1/2) K (x₁ + x₂)² [Equation 2]
Equating Equation 1 and Equation 2
F(x₁ + x₂) = (1/2) K (x₁ + x₂)²
(x₁ + x₂) = 2F / K
Since , net external force on two blocks is zero thus , same force act on opposite direction.
Therefore ,
m₁x₁ = m₂x₂ [Equation 3]
(x₁ + x₂) = 2F / K (as shown above)
x₂ = m₁ x 1 /m₂
Substituting , ( m₁ x 1 / m₂ ) + x₁ = 2F / K
x₁( 1 + m₁/m₂) = 2F / K
x₁ = 2F m₂ / K (m₁ + m₂)
Similarly , x₂ = 2F m₁ / K ( m₁ + m₂)
Hope it helps!
