Combining Conservation Laws. A 4.75 kg} chunk of ice is sliding at 10.0 m/s} on

Question

Combining Conservation Laws.

A 4.75 kg} chunk of ice is sliding at 10.0 m/s} on the floor of an ice-covered valley when it collides with and sticks to another 4.75 kg} chunk of ice that is initially at rest. (See the figure below (Figure 1) .) Since the valley is icy, there is no friction.

****FIGURE 1 GO TO THIS LINK: http://session.masteringphysics.com/problemAsset/1263256/5/yg.8.37.jpg

Part A
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.)
H = m

Explanation / Answer

first momentum 4.75*10= 2*4.75*v so they will have a velocity of v=5 conservation of energy 1/2 mv^2 = mgh 0.5*5^2 = 9.81*h h=0.5*25/9.81=1.27 m

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