A 4.95kg chunk of ice is sliding at 10.5m/s on the floor of an ice-covered valle

Question

A 4.95kg chunk of ice is sliding at 10.5m/s on the floor of an ice-covered valley when it collides with and sticks to another 4.95kg chunk of ice that is initially at rest. (See the figure below (Figure 1) .) Since the valley is icy, there is no friction.

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.)

---Please show work and not just equations. Thanks

Explanation / Answer

using conservation of momentum concept

initial total momentum = final total momentum

mu + zero = (m+m)*V

V = u/2 = 10.5/2 = 5.25 m/s

when it reaches the heighest point its kinetic energy is zero , it only has potential energy

using conservation of energy

.5(2m)V^2 = 2m*gH

H = 1.405 m

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