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Question

A 5 kg block of lead (c = 128 J/(kg °C)) is at rest on a spring with a constant of 150 N/m. A 2 kg block of lead is thrown on top of the 5 kg block from a height of 2.40m above the block and an initial speed of 5 m/s, and they stick together as they move downwards. Assume there is no air resistance involved, and that both blocks are at a temperature of 25 °C before colliding.

a) How far is the spring compressed from its natural length before the collision happens, in m.

b) How fast is the 2 kg block of lead moving just before colliding with the 5 kg block, in m/s?

c) How fast are the two blocks of lead moving together just before they begin to compress the spring further, in m/s?

d) What is the change in temperature of the lead blocks right after colliding, in °C? Assume all lost energy went into this process.

Answer 1

Each of these problems involve different scenarios with different relevant equations:

a) Here, we are looking at a compression of a spring, which is governed by Hooke's Law: The Restorative Force = - Spring Constant · Displacement from Equilibrium: F = -kx

What is causing the compression is the force of gravity from the 5 kg lead block: mg

Therefore, mg = kx (taking the absolute value), so

x = mg/k = (5 kg ⋅ 9.8 m/s²) / 150 N/m = 0.327 m

b) Before the collision, the 2 kg block is at some unknown velocity due to its falling. So, we first need to solve for that. The easiest way is using the conservation of energy. At the top, the 2 kg block as an initial speed of 5 m/s (so it has some initial kinetic energy) and is at a height of 2.4 m (so it has some potential energy). Even if there is friction / air resistance, we can always set the initial thermal energy to zero for these problems (just like we can usually set the initial x position to equal zero). Therefore:

Total Energy at the top = Kinetic + Potential + Thermal = 1/2 m v² + mgh + 0 = 1/2 · (2 kg) · (5 m/s)² + 2 kg · 9.8 m/s² · 2.4 m = 72.04 J

At the bottom, since the is no air resistance and the final height is zero, all of that energy is kinetic.

Total Energy at the bottom = Kinetic at the bottom = 1/2 m v²

v = sqrt (Total E · 2 / m) = sqrt(72.04 J · 2 / 2 kg) = 8.51 m/s

c) Now back to the collision. Collisions involve the conservation of momentum: Total initial momentum = Total final momentum, where momentum is defined as the mass of the object times its velocity: p = mv.

Before, only that 2 kg block is moving, so the total initial momentum = 2 kg · 8.51 m/s = 17.02 kg · m/s

After the collision, they stick together and function as a single object with their combined mass·, so the total momentum = (m_A + m_B) v_F

v_F = Total p / (m_A + m_{B) =} 17.02 kg · m/s / (2 kg + 5 kg) = 2.43 m/s

d) This collision is inelastic since the objects stick together, so some of the initial kinetic energy gets transferred to heat during the collision, and we already found the initial kinetic energy to be the 72.04 J from before. After the collision, the kinetic energy of the two blocks is 1/2 (m_A + m_B) v_F² = 1/2 · ·(2 kg + 5 kg) · (2.43 m/s)² = 20.67 J. Therefore, the remaining 51.37 J was converted to heat during this process.

The heat flow resulting in a temperature change is equal to the mass (7 kg now) times the specific heat of the object and its phase (128 J / kg °C) times the change in temperature: Q = mC ΔT

Therefore, ΔT = Q / mC = 51.37 J / (7 kg · 128 J / kg °C) = 0.057 °C

The unfortunate thing about this set of problems is that a minor mistake in part b affects everything that follows! The important thing, however, is understanding the steps and physics involved