Q. A car of mass 1000 kg is going towards east at 25.0 m/s. It hits a slow moving truck (moving at speed 10.0 m/s also going towards east) of mass 2500 kg from back.
(a) If after the collision the truck is moving at 15.0 m/s towards east, what is the velocity of the car?
(b) If the collision between the car and the truck is perfectly elastic, find their velocities after the collision.
(c) Find the percentage loss of kinetic energy of the vehicles due to the collision in both cases.
Car of 1000kg going east at 25 m/s hits a kid of 2500 kg going east at 10 m/s. what is the velocity of car after collision if kid moved 15ms?
2020-05-12 1:14 pm
回答 (2)
2020-05-12 3:22 pm
Ic = 1,000*25 = 25,000 kg*m/sec
It = 2,500*10 = 25,000 kg*m/sec
I = Ic+It = 50,000 kg*m/sec
KEc = 25^2*1000/2 = 312,500 joule
KEt = 10^2*2500 = 250,000 joule
KE = 562,500 joule
a)
Vt' = 15 m/sec
It' = 2,500*15 = 37,500 kg*m/sec
Vc' = I-It')/mc = (50,000-37,500)/1000 = 12,5 m/sec
KE' = 500*12.5^2+15^2*1,250 = 359,375 joule
k = 100*(562,500-359,375)/562,500 = 36.11 %
b)
prior impact
I = 50,000 kg*m/sec
KEc = 25^2*1000/2 = 312,500 joule
KEt = 10^2*2500 = 250,000 joule
KE = 562,500 joule
after impact
both momentum and kinetic energy are conserved
mc*Vc' + mt*Vt' = 50,000
mc*Vc'^2+mt*Vt'^2 = 1,125,000 joule
Vc' = (50,000-mt*Vt')/1000
1000*((50,000-2500*Vt')/1000)^2+mt*Vt'^2 = 1,125,000
1000*(2,500+6.25Vt'^2-250Vt')+2500*Vt'^2 = 1,125,000
2,500,000+6,250Vt'^2-250,000Vt'+2,500*Vt'^2 = 1,125,000
8750Vt'^2-250,000Vt'+1375,000 = 0
8.75Vt'^2-250Vt'+1375 = 0
Vt' = (250 ± √ 250^2-8.75*4*1375)/17.5 = 21.137 ; 7.435 m/sec
Vt' must be higher than Vt, therefore 21.137 value will suit !!
50,000-21.137*2,500 = -2.842 kg*m/sec
Vc' = -2.842/1000 = -2.842 m/sec
KE'' = -2,842^2*1000+21,137^2*2500 = 1,125,000 joule... no loss of energy at all
It = 2,500*10 = 25,000 kg*m/sec
I = Ic+It = 50,000 kg*m/sec
KEc = 25^2*1000/2 = 312,500 joule
KEt = 10^2*2500 = 250,000 joule
KE = 562,500 joule
a)
Vt' = 15 m/sec
It' = 2,500*15 = 37,500 kg*m/sec
Vc' = I-It')/mc = (50,000-37,500)/1000 = 12,5 m/sec
KE' = 500*12.5^2+15^2*1,250 = 359,375 joule
k = 100*(562,500-359,375)/562,500 = 36.11 %
b)
prior impact
I = 50,000 kg*m/sec
KEc = 25^2*1000/2 = 312,500 joule
KEt = 10^2*2500 = 250,000 joule
KE = 562,500 joule
after impact
both momentum and kinetic energy are conserved
mc*Vc' + mt*Vt' = 50,000
mc*Vc'^2+mt*Vt'^2 = 1,125,000 joule
Vc' = (50,000-mt*Vt')/1000
1000*((50,000-2500*Vt')/1000)^2+mt*Vt'^2 = 1,125,000
1000*(2,500+6.25Vt'^2-250Vt')+2500*Vt'^2 = 1,125,000
2,500,000+6,250Vt'^2-250,000Vt'+2,500*Vt'^2 = 1,125,000
8750Vt'^2-250,000Vt'+1375,000 = 0
8.75Vt'^2-250Vt'+1375 = 0
Vt' = (250 ± √ 250^2-8.75*4*1375)/17.5 = 21.137 ; 7.435 m/sec
Vt' must be higher than Vt, therefore 21.137 value will suit !!
50,000-21.137*2,500 = -2.842 kg*m/sec
Vc' = -2.842/1000 = -2.842 m/sec
KE'' = -2,842^2*1000+21,137^2*2500 = 1,125,000 joule... no loss of energy at all
2020-05-12 1:15 pm
If you always get someone else to do your homework you will never understand anything
收錄日期: 2021-04-24 07:50:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200512051427AAUmyUq