1)On a horizontal frictionless surface, a block of mass M is sliding along on co

Question

1)On a horizontal frictionless surface, a block of mass M is sliding along on constant velocity v. A block of mass m lands on top of the original block. What is the velocity of the new system (M+m)?



2)Consider a spring handing vertically from a ceiling, attached to a mass m of 2 kg. The spring constant, k, is 30 N/m. The spring is compressed 0.8m and then released. How far does the spring travel downwards before it reverses direction? (hint: consider both gravitational and elastic potential energies)

Explanation / Answer

1) Conservation of Momentum : M*v = (M+m)*v' v' = Mv / (M+m) 2) Initial SPE = 1/2*30*0.8^2 = 9.6 J Initial GPE = 2*9.81*0.8 = 15.696 J Let Final elongation = x Final SPE = 15x^2 Final GPE = -2*9.81*(x) So, 15x^2 - 19.62x = 25.296 x = 2.1 m So spring travels (2.1+0.8) = 2.9 m downwards before it reverses direction

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

---

---

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