Question

1. A block of mass m = 0.23 kg slides with initial velocity v₀ = 1.4 m/s along a frictionless surface (as in the figure). The block next slides through a region of length d = 0.25 m with kinetic friction. After passing through the friction region, the block slides up a curved ramp until momentarily coming to rest at a height h above the level surface. If the kinetic friction coefficient μ k = 0.204, what height h does the block reach? Express the answer in meters but enter only the numerical part in the box.

2. Two blocks (masses m₁ and m₂) are on the double ramp (as shown). The region of length L is horizontal. The masses are released at the same time from points A and B (at heights h₁ = 0.4 m and h₂ = 0.29 m, respectively). The collision of the two masses (in the horizontal section) is completely inelastic. What is the ratio of the two masses (m₁/m₂) if the combined mass climbs to a height h₃ = 0.057 after the collision? Your answer will be unitless (a simple ratio entered as a decimal number).

Answer 1

Initial kinetic energy = (1/2)*0.23*(1.4)2 = 0.2254 J

Work by friction = -(0.204*0.23*9.8)*0.25 = 0.1150 J

Final Mechanical energy = 0.2254 - 0.1150 = 0.1104 J

0.1104 = 0.23*9.8*h ; hence h = 0.049 metres