Problem 8.19: Combining Conservation Laws. Part A A 4 70 kg chunk of ice is slid

Question

Problem 8.19: Combining Conservation Laws. Part A A 4 70 kg chunk of ice is sliding at 15 0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 4 70 kg chunk of ice that is initially at rest (See the figure below (Figure 1)) Since the valey is icy, there is no friction After the collision, how high above the valley floor will the combined chunks go? (Hint Break this problem into two parts-the collision and the behavior after the collision-and apply the appropriate conservation law to each part) Faure 1 of 1 Submit My Answers Give Up Continue

Explanation / Answer

First conserve momentum, then conserve energy.
(i) initial p = final p
4.70 kg * 15 m/s + 0 = (4.70 + 4.70)kg * v
v = 7.5 m/s post-collision velocity

(ii) The combined KE gets converted into PE:
½mv² = mgh multiply through by 2/m
v² = 2gh
h = v² / 2g

= (7.5 m/s)² / 19.6m/s² = 2.87 m

since there is no friction, this height is independent of the path taken as well as the mass of the ice.

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