10/16/2017 Suggested 9.3 In the figure below, two boxes are sliding to the right

Question

10/16/2017 Suggested 9.3 In the figure below, two boxes are sliding to the right on a horizontal frictionless surface. Box 1 has a mass of 1 kg and a speed of 4 m/s, while box 2 has a mass o kg and a speed of 2.5 m/s. Eventually, box 1 catches up to box 2 and experiences a completely inelastic collision with box 2. f 4 a. What is the momentum of the two box system after the collision? (14 kg m/s) b. What is the speed of the two blocks after the collision? (2.8 m/s) c. is the total kinetic energy of the two blocks after the collision greater than, less than, or equal to the total kinetic energy before the collision? Less than. m1 = 1 kg m1 = 4 kg 4 m/s 2.5 m/s Suggested #3

Explanation / Answer

Here ,

m1 = 1 Kg

u1 = 4 m/s

m2 = 4 Kg

u2 = 2.5 m/s

a) as the momentum of the collision will be conserved

momentum of the two box system = m1 * u1 + m2 * u2

momentum of the two box system = 1 * 4 + 4 * 2.5

momentum of the two box system = 14 Kg.m/s

b)

let the speed after collision is v

(m1 + m2) * v = 14

(1 + 4) * v = 14

v = 2.8 m/s

the speed of two blocks is 2.8 m/s

c)

as the collision is inelastic collision , the kinetic energy will be lost

the final kinetic energy less than the initial kinetic energy

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