View Full Version : Car Speed calculation at the time of accident
naval_kc
Jan 5, 2009, 09:33 AM
Vehicle No. 1 (going from North to South) hit Vehicle No. 2 (going from East to West) at roads intersection. After the collision Vehicle 1 dragged Vehicle 2 around 70 m in the direction from North to South. Weight of Vehicle No. 1 is 2.5 Tons and of Vehicle No. 2 is 2.2 Tons. Given the coefficient of friction (µ) 0.8. Calculate the speed of Vehicle No. 1 at the time of collision?
Case (I): When the vehicle no. 2 was stopped at the time of collision
Case (II): When the vehicle no. 2 was moving at a speed of 60 km/hr.
harum
Jan 9, 2009, 07:06 PM
Use the momentum conservation law to find the speed of both cars right after collision. And remember that the car 2 had no momentum along the direction North to South (you can call this direction axis Y) in either case. You also have to assume that the cars move together as one object of combined mass after collision. Please note that the sum of projections of the momenta of the cars is the same right before and right after the collision. After collision the friction works to reduce kinetic energy of two cars 0.5*(M1+M2)*V^2 into heat, here V is the speed of the cars right after collision. Let us know whether this helps or not.
Rbeuttler
Jun 12, 2010, 05:25 PM
m1 skided 37 feet beforecolsion
Rbeuttler
Jun 12, 2010, 05:27 PM
My car skided 37 feet before colison
Rbeuttler
Jun 12, 2010, 05:37 PM
m1 skided 37 feet before colsion
m1 stopped 5 feet from inmpaked
Rbeuttler
Jun 12, 2010, 05:42 PM
Use the momentum conservation law to find the speed of both cars right after collision. And remember that the car 2 had no momentum along the direction North to South (you can call this direction axis Y) in either case. You also have to assume that the cars move together as one object of combined mass after collision. Please note that the sum of projections of the momenta of the cars is the same right before and right after the collision. After collision the friction works to reduce kinetic energy of two cars 0.5*(M1+M2)*V^2 into heat, here V is the speed of the cars right after collision. Let us know whether this helps or not.
I do not understand calculus
Rbeuttler
Jun 12, 2010, 05:43 PM
I do not understand calculus
Rbeuttler
Jun 12, 2010, 05:48 PM
m1 skided 37 feet beforecolsioncar #2 was moving at a speed of 10mi per hr and car # 2 moved abour 3 feet after impaced
Rbeuttler
Jun 12, 2010, 05:49 PM
What was #1 speer
Fr_Chuck
Jun 13, 2010, 12:58 PM
Thank you for using ask me help desk for your questions.
This is a obvious home work question, since it did not ask about issues that really happen in actual wrecks, type of payment, weather conditions, and many other factors not addressed.
Even then if it is a obvious home work, we don't do it,
Give us your idea of how you would do it, and someone where may be glad to help you.
Also stop asking similar or the same questions and deal with your question
phone3
Apr 4, 2011, 01:28 PM
Would this be considered a minor accident and what are the chances of injury resulting from the above