A 50 kg mass moving to the right at 8 m s−1 collides with a 25 kg mass moving to the right at 3 m s−1. After the collision, the velocities of the masses are v1 and v2, where v2 > v1 and 15% of the original kinetic energy is lost during the collision. Assume that no external forces are acting.
(a) Show that v2 = 19 − 2v1.
(b) Show that 3v 2 1 − 38v1 + 100 ≈ 0 is a good approximation of the equation for v1.
(c) Calculate v1 and v2.
(d) What happened to the 15% lost energy?
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Expert's answer
2021-05-11T09:07:18-0400
before collision
m1=50;u1=8
m2=25;u2=3
m1∗u1+m2∗u2=475
Ek1=2m1u12+2m2u22=1712.5
after collision
m1=50;v1
m2=25;v2
m1∗v1+m2∗v2=m1∗u1+m2∗u2=475(1)
Ek=2m1v12+2m2v22
Ek=Ek1−0.15∗Ek1=1455.625
Ek=2m1v12+2m2v22=1455.625(2)
(a) from the equation(1)
m1∗v1+m2∗v2=475
50∗v1+25∗v2=475
2∗v1+v2=19
v2=19−2∗v1(3)
(b) from the equation(2)
2m1v12+2m2v22=1455.625
m1v12+m2v22=2911.25
50v12+25v22=2911.25
2v12+v22=116.45
from the equation(3)
2v12+(19−2v1)2=116.45
6v12−76v1+244.55=0
3v12−38v1+122.75=0
3v12−38v1+100≈0(4)
(c) from the equation(4)
3v12−38v1+100=0
v1=3.8;v2=19−2∗v1=11.4
Decision v1′=8.9 is not taken by the condition of the task v2>v1
Answer: v1=3.8;v2=11.4
(d) Part of the translational kinetic energy is transferred into the internal energyof the colliding bodies.
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