Question

A4.80 kg chunk of ice is sliding at 13.0 m/s on the floor of an ice-covered valley when it collides with and sticks to another 4.80 kg chunk of ice that is initially at rest. (See (Figure 1).) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. Part A How fast are they moving when they reach the valley floor again? Express your answer in meters per second. View Available Hint(s) V AL¢ a e ? m /s Submit Provide Feedback Figure 1 of 1

Answer 1

As there is no frictional force total energia is conserved. Hence after collision if the sticked up ice has velocity is v , then after clinding up and then down the hillside, it has the same Velocity v. Now, we have to determine r. From the conservation of momentum, 4.80813 + or v= 4.8000 = (4.8+ 4.8) v 4.8 X 13 = 16.5 m/s) Ang 9.6