Problem 14.32 says: determine the energy loss as a result of the series of colli
ID: 1770704 • Letter: P
Question
Problem 14.32 says: determine the energy loss as a result of the series of collisions described in problem 14.7 14.7 At an amusement,park there are 200-kg·bumper cars A, B, and o ,and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectivel Car A is moving to the right with a velocity vA= 2 m/s and carc has a velocity vB 1.5 m/s to the left, but car B is initially at rest The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming (a) cars 4 and C hit car B at the same time, (b) car A hits car B before car C does C: Fig. P14.7 and P14.8
Explanation / Answer
14.7. Case (a)
A and C hit B at the same time
Ua = 2m/s
Ub = 0 m/s
Uc = -1.5m/s
For A and B
As coefficient of restitution is 0.8 so equation be like
Vb - Va = 0.8*(Ua-Ub)
Vb - Va = 1.6 ...(1)
For B and C
Vc - Vb = 0.8(Ub-Uc)
Vc - Vb = 1.2 ...(2)
Now Conservation of momentum
240*Ua + 260* Ub +235* Uc = 240*Va + 260 *Vb + 235 *Vc
240*Va + 260*Vb + 235*Vc = 127.5 ...(3)
On solving we get
Va = -1.287 m/s
Vb = 0.312 m/s
Vc = 1.512 m/s [Ans (a)]
Case (b) A hits B before C does
For Collision of A and B
Vb - Va = 1.6
240*Ua = 240*Va +260Vb = 480
So Va = 0.128 m/s
Vb =1.728 m/s
Now for collision of B and C
Vb2 - Vc = 0.8*(Uc-Vb)
Vb2 - Vc = -2.582
260*Vb + 235 * Uc = 260*Vb2 + 235 * Vc
so 260*Vb2 + 235 * Vc = 96.78
So Vb2 = -1.03 m/s
Vc = 1.55m/s
Now A will again hit B with velocities Va=0.128 and Vb2 = -1.03
so Vb3 - Va2 = 0.8(Va-Vb2) = 0.9264
240*Va + 260*Vb2 = 240*Va2 + 260*Vb3 = -237.08
So Va2 = -0.955
Vb3 = -0.029
Final velocities of A B and C after all collisions are
Vaf = -0.955 m/s
Vbf = -0.029 m/s
Vcf = 1.55 m/s [Ans(b)]
14.32
Change in energy of A= 0.5 ma (22-0.9552)= 370.557 J
Change of energy in B= 0.5 mb(0-0.0292)= -0.11J
Change of energy in C= 0.5 mc (1.52-1.552)= -17.92 J
Net loss of energy =370.557-0.11-17.92= 352.527 J
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