Two gliders are set in motion on an air track. A spring of force constant k is a

Question

Two gliders are set in motion on an air track. A spring of force constant k is attached to the near side of one glider. The first glider of mass m1 has velocity vector v 1 and the second glider of mass m2 moves more slowly, with velocity vector v 2, as in the figure shown below. When m1 collides with the spring attached to m2 and compresses the spring to its maximum compression xmax, the velocity of the gliders is vector v . Find each of the following in terms of vector v 1, vector v 2, m1, m2 and k. (Use v_1 for vector v 1, v_2 for vector v 2, m_1 for m1, m_2 for m2 and k for k as necessary.)
(a) the v at maximum compression
(b) the maximum compression xmax
(c) the velocity of each glider after m1 has lost contact with the spring

Explanation / Answer

a)By conservation of momentum,
(m1 + m2)*vmax = m1 v1 + m2 v2
vmax = (m1 v1 + m2 v2)/(m1 + m2)

b)By conservation of energy,

(1/2)(m1 + m2) vmax^2 + (1/2)k xmax^2 = (1/2)m1 v1^2 + (1/2)m2 v2^2

(1/2)(m1 + m2)(m1 v1 + m2 v2)^2/(m1 + m2)^2 + (1/2)k xmax^2 = (1/2)m1 v1^2 + (1/2)m2 v2^2

Multiplying by 2,
(m1 + m2)(m1 v1 + m2 v2)^2/(m1 + m2)^2 + k xmax^2 = m1 v1^2 + m2 v2^2

(m1 + m2)(m1 v1 + m2 v2)^2/(m1 + m2)^2 + k xmax^2 = m1 v1^2 + m2 v2^2

(m1 v1 + m2 v2)^2/(m1 + m2) + k xmax^2 = m1 v1^2 + m2 v2^2

k xmax^2 = m1 v1^2 + m2 v2^2 - (m1 v1 + m2 v2)^2/(m1 + m2)

xmax^2 = [m1 v1^2 + m2 v2^2 - (m1 v1 + m2 v2)^2/(m1 + m2)]/k

xmax = sqrt{[m1 v1^2 + m2 v2^2 - (m1 v1 + m2 v2)^2/(m1 + m2)]/k}

c) Let velocity of 1st glider = v1f,
of second glider = v2f

(1/2)m1 v1f^2 - (1/2)m1 vmax^2 = (1/2k xmax^2
(1/2)m2 v2f^2 - (1/2)m2 vmax^2 = (1/2k xmax^2

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.