A 5 kg block with a speed of 3 m/s collides with a 10 kg block that has a speed

Question

A 5 kg block with a speed of 3 m/s collides with a 10 kg block that has a speed of 2 m/s in the same direction. After the collision, the 10 kg block travels in the original direction with a speed of 2.5 m/s. A) What is the velocity of the 5 kg block imediately after the collision? B) By how much does the total kinetic energy of the system of two blocks change because of the collision? C) Suppose, instead, that the 10 kg block ends up with a speed of 4 m/s. What then is the change in the total kinetic energy? D) Account for the result you obtained in (C).

Explanation / Answer

let the speed be v conservation of momentum 5 * 3 + 10* 2 = 10 * 2.5 + 5* v 5v = 10 v = 2 m/s ----answer b) initial KE 0.5 * 5 * 9 + 0.5 * 10 * 4 = 42.5J final KE 0.5 * 10 * 6.25 + 0.5* 5 * 4 = 41.25 J loss of KE = 41.25 - 42.5 = 1.25 J ---ans c) 5 * 3 + 10* 2 = 10 * 4 + 5* v1 v1 =-1 final KE 0.5 * 10 * 16 + 0.5* 5 * 1 = 82.5 J change of KE = 82.5 -42.5 = 40 J ---ans d) not possible to add on KE after impact unless 10 kg is powered by itself. ---answer Source(s): my brain (Prof TBT)

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