Two blocks are free to slide along the frictionless wooden track shown below. Th

Question

Two blocks are free to slide along the frictionless wooden track shown below. The block of mass m_1 = 4.98 kg is released from the position shown, at height h = 5.00 m above the flat part of the track. Protruding from its front end is the north pole of a strong magnet, which repels the north pole of an identical magnet embedded in the back end of the block of mass m_2 = 9.60 kg, initially at rest. The two blocks never touch. Calculate the maximum height to which m_1 rises after the elastic collision.

Explanation / Answer

Given :-

m1 = 4.98 kg

m2 = 9.60 kg

elastic collision ( restitutive coefficient e = 1)

principle of conservation energy

consider at free body of mass m1

Energy at A = Energy at B

m1gh = ½ m1v1²

v1² = 2gh = 2(9.8)(5)

v1 = 9.8995 m/s

e = -(v1' - v2')/(v1 - v2) = 1

e = -(v1' - v2')/(9.8995 - 0) = 1

v2' = 9.8995 + v1'

equation of conserve momentum

m1 v1 + m2 v2 = m1 v1' + m2 v2'

(4.98) v1 + (9.60) (0) = (4.98) v1' + (9.60) v2'

remember v2' = 9.8995 + v1'

(4.98) (9.8995) + (9.60) (0) = (4.98) v1' + (9.60) (9.8995 + v1')

49.2995 = 14.58v1' + 95.0352

v1' = -3.1369 m/s

again, conserve of energy

mgh' = ½ m(v1')²

h' = (v1')²/2g

h' = (-3.1369)²/{2(9.8)}

h' = 0.5 m

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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