A 5.00{\ m kg} chunk of ice is sliding at 10.0{\ m m}/{\ m s} on the floor of an

Question

A 5.00{ m kg} chunk of ice is sliding at 10.0{ m m}/{ m s} on the floor of an ice-covered valley when it collides with and sticks to another 5.00{ m kg} chunk of ice that is initially at rest. (See the figure below (Figure 1) .) Since the valley is icy, there is no friction. 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.)

Explanation / Answer

the momentum is conserved in the collision: m1v1 + m2v2 = (m1+m2)*v3 with m1,m2 = masses of particle 1, respectively 2 v1, v2 = velocities of particles 1, respectively 2 v = velocity after collision 5*10 + 5*0 = (5 + 5)*v3 v3 = 5*10/10 = 5 m/s the kinetic energy of the united particles is 1/2*(m1+m2)*v^2, which is equal to the potential energy after moving up a height h: 1/2*10*5^2 = mgh 125 = 10*9.81*h h = 125/(98.1) h = 1.27 m

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