Collision in Two Dimensions
2007-08-11 11:06 am
The police are investigating an accident. They have determined that the mass of car A is 2275kg and the mass of car B is 1525kg. From the skid marks and date for the friction btw tires and concerte, the police determined that the cars, when they were locked together, had a velocity of 31km/h at angle of 43 degrees north of the eastbound street. If the speed limit was 35km/h on both streets, should one or both cars be ticketed for speeding?
回答 (2)
2007-08-20 6:33 pm
✔ 最佳答案
圖片參考：http://hk.geocities.com/namen1997/P0008.jpg
問題的角度不清楚，可能會有兩種情況，第一種如上圖所示：
在x軸
mAu1 = (mA + mB)v cosθ
2275u1 = (2275 + 1525) x 31 cos43o
u1 = 37.9km/h
在y軸
mBu2 = (mA + mB)v sinθ
1525u2 = (2275 + 1525) x 31 sin43o
u2 = 52.7km/h
則兩車都超速。
另一種情況，如上圖但A和B的位置對掉，則
mAu1 = (mA + mB)v sinθ
2275u1 = (2275 + 1525) x 31 sin43o
u1 = 35.3km/h
在y軸
mBu2 = (mA + mB)v cosθ
1525u2 = (2275 + 1525) x 31 cos43o
u2 = 56.5km/h
則兩車都超速。
2007-08-13 3:01 am
Let u1 and u2 be the velocities of cars A and B respectively.
By resoloving the velocity respectively in directions along the line of impact (i.e. 43 degrees north of the eastbound street) and perpendicular to the line of impact, we have, by conservation of momentum:
2275(u1).cos(43) - 1525(u2)cos(43) = (2275+1525)(31)
and 2275.(u1).sin(43) = 1525.(u2).sin(43)
solve for u1 and u2
By resoloving the velocity respectively in directions along the line of impact (i.e. 43 degrees north of the eastbound street) and perpendicular to the line of impact, we have, by conservation of momentum:
2275(u1).cos(43) - 1525(u2)cos(43) = (2275+1525)(31)
and 2275.(u1).sin(43) = 1525.(u2).sin(43)
solve for u1 and u2
收錄日期: 2021-04-13 01:02:18
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070811000051KK00570