A BMW of mass 1.2 103 kg is traveling at 41 m/s. It approaches a 1.2 103 kg Volkswagen going 26 m/s in the same direction and strikes it in the rear. Neither driver applies the brakes. Neglect the relatively small frictional forces on the cars due to the road and due to air resistance. If the collision slows the BMW down to 32 m/s, what is the speed of the VW after the collision?
2006-10-20 13:22:31 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Physics
the mass of the car is 1.2*10^3
2006-10-20 13:42:43 · update #1
This is a conservation of momentum question. The total momentum of both cars before the prang is m1*41(BMW)+m1*26(VW) = m1*67 because they have the same weight .
After the collision the momentum is the same and = m1(32+v2 of VW)
v2 of VW= 35m/sec
2006-10-20 14:36:27 · answer #1 · answered by sydney m 2 · 0⤊ 0⤋
What in the world is 1.2 103 kg?
Is this like Stuart Little's car ?
2006-10-20 20:33:00 · answer #2 · answered by Ren Hoek 5 · 0⤊ 0⤋
This problem deals with the conservation of momentum. In this case, the momentum of both cars before collision is equal to their momentum after collision.
Before collision:
Momentum of BMW=m*41
Momentum of Volks= m*26
Total momentum of both cars=41m+26m
After collision:
Momentum of BMW:=m*32
Momentum of Volks =m*v
Total momentum of both cars=32m+m*v
Total Momentum before collision = Total
Momentum after collision
41m+26m=32m+mv. Do you agree that m cancels out?
If so,
41+26=32+v
41+26-32=v
35=v
v=35m/s
2006-10-22 11:05:42 · answer #3 · answered by tul b 3 · 0⤊ 0⤋