Question

whicle The driver forgot to tie lerates forward at 7. A clay block of mass m moving to the right with speed V, strikes a second clay block of mass 2m moving to the left with the speed vd/3. The two clay blocks stick together after the collision. There are no outside forces acting on the blocks. What is the final velocity of the two blocks? A) m vo to right B) 1/3 v. to left C) (1/2) m V, to right D) 3/4 m v. to left E) (1/9) v. to right 8. In problem 7, calculate the percentage of the original kinetic energy that is converted into heat: A) 39% B) 96.7% C) 64% D) 37 % E None of these

Answer 1

Let - us sign use 't' sign for velocity towards towards for velocity deft. right and mi = m -۷ ملا m2 = V2 = 2m - Vo 13 Consenving momentum before and after_collision mivu + m2 V2 = (mi + m2) v' sticking? - lvie final velocity_of_2_blocks after : mv. + (2m x -vo) = (m + 2m) v - mv. - 2mv. - amvi & mv = 3mvi 3 = v' = + Vo final velocity = vo to right option E]

-8. Original._ K.E. Converted into heat - = Change in K.E. before and after collision. = mvi? + ļ my w32 1-m. + m2) 1:27 K.E. col - system collision K.E. cy after system collision. -{mve? + (1x2mxfiy} - {}x (420) 4 (4) ? = 1 mx ² + mv 2 ur na =.mw? [ + ģista] - Initial K.E. before collision = 1 m, v 2 + 1 m2 V₂2 Change in KE Snitial KE X100

zemvo - X 100 1 mv. ² X100 11 (96.969 % Coption o]
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