HomeStudent: bondz Class Management |Help HW 19 Begin Date: 8/9:2017 12:00:00 AM

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HomeStudent: bondz Class Management |Help HW 19 Begin Date: 8/9:2017 12:00:00 AM --Due Date: 11/7/2017 11:59:00 PM End Date: 11/10 2017 11:59:00 PM (14%) Problem 3: A body of mass 2.3 kg and initial speed of39 m/s collides with an initially resting body of mass 17 kg. The bodies stick together after the collision and continue moving (in the direction of the initially moving body). The motion takes place on a horizontal flat surface with negligible friction. 33% Part (a) Find the speed of the combined body immediately after the collision in meters per second Grade Summary Deductions Potential 0% 100% sin) cotan() Submissions Attempts remaining: 4 cos asin() atanacotansinhO cosh tanh) cotanh) tan( ) ( acos % per attempt) detailed view END DegreesRadians Submit Hint I give up! Hints: 0% deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback. 33% Part (b) This collision is very inelastic. Calculate the loss of kinetic energy, in Joules, by subtracting the kinetic energy of the combined body immediately after the collision from the kinetic energy of the initially moving body immediately before the collision. 33% Part (c) Find the increase in the internal energy, in joules, of the two bodies during the collision. All content 2017 Expert TA. LLC

Explanation / Answer

Part (a)

m1 = 2.3 kg, u1 = 39 m/s

m2 = 17 kg, u2 = 0

apply conservation of momentum -

m1u1 + m2u2 = (m1+m2)v

put the values -

2.3*39 + 17*0 = (2.3+17)*v

=> v = (2.3*39) / 19.3 = 4.65 m/s

Part (b) -

Initial kinetic energy, Ki = (1/2)m1u1^2 = 0.5*2.3*39^2 = 1749.15 Joule

Final kinetic energy, Kf = (1/2)(m1+m2)v^2 = 0.5*19.3*4.65^2 = 208.66 J.

So, loss in kinetic energy = Ki - Kf = 1749.15 - 208.66 = 1540.49 J

Part (c)

Increase of internal energy = loss in the kinetic energy = 1540.49 J

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