Q12: A chunk of ice of mass m is sliding at speed v on the floor of an ice-cover

Question

Q12: A chunk of ice of mass m is sliding at speed v on the floor of an ice-covered valley when it collides with and sticks to a chunk of ice which is twice as massive, 2m, initially at rest. (See the figure below). Assume that there is no friction. The answers to parts (a). (b). (c) and Extra Credit below should be expressed in terms if m, v and g only. FFR 2mm (a) [4 pts] Right after the collision, what is the speed of the two chunks sticking together? (b) [4 pts] What is the total energy of the system right after the collision (right before the combined chunks start going up the slope)? Assume that the bottom of the valley is zero potential energy. (e) 18 pts] After the collision, how high above the valley floor will the combined chunks go? EXTRA CREDIT [4 pts] Has any energy been lost in the collision? If so, how much?

Explanation / Answer

a )Law of conservation of momentum

Let velocity of combination is v'

mv+0=(m+2m)v'

v'=v/3

b) Total energy =1/2×3m×v^2/9=mv^2/6 J

C )Conservation of energy

3mgh=mv^2/6

h=v^2/18g

d) collision is inelastic so there wil be loss of energy

K.E final =1/2×3m×v^2/9=mv^2/6

K.E initial =1/2mv^2

Change in energy =mv^2 (1/6-1/2)=-2/3mv^2

Negative sign here shows that there wil be loss of energy

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