A 3.00-kg chunk of ice is sliding at 10.0m/s on the floor of an ice-covered vall

Question

A 3.00-kg chunk of ice is sliding at 10.0m/s on the floor of an ice-covered valley when it collides with and sticks to another 7.00-kg chunk of ice that is initially at rest as in figure below. Since the valley is icy, there is no friction. What is the speed of the combined chunk of ice after collision? Find the total kinetic energy before and after the collision. What percentage of of the kinetic energy has dissipated into heat in this collision? How high above the valley floor will the combined chunks go?

Explanation / Answer

a) 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

3*10 + 7*0 = (3+7)*v3
v3 = 3 m/s

b) Initial kinectic energy = 1/2m1v1^2 + 1/2m2v2^2 = 1/2*3*10*10 +0 = 150J

Final kinetci energy = 1/2 (m1+m2)V3^2 = 1/2 * 10*3*3 =45 J

energy lost = 105 J

c) 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:
45 = mgh
45 = 10*9.81*h
h = 2.18 m

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