Asked by Corey
As a van is turning left in an intersection, it crashes into an oncoming car. The van has a mass of 4000 kg and was traveling 55° East of North at 12 km/h before the crash. The car has a mass of 2500 kg and was traveling West at 55 km/h before the crash. After the crash, the van bounces off at 36 km/h to the west (ignore any rotational motion of the car and van). What is the velocity of the car immediately after the crash?
Answers
Answered by
oobleck
Conserve momentum
4000<36,0> + 2500<vx,vy> = 4000<12sin55°,12cos55°> + 2500<-55,0>
vx = (4000*12sin55° - 2500*55 - 4000*36)/2500
vy = (4000*12cos55°)/2500
4000<36,0> + 2500<vx,vy> = 4000<12sin55°,12cos55°> + 2500<-55,0>
vx = (4000*12sin55° - 2500*55 - 4000*36)/2500
vy = (4000*12cos55°)/2500
Answered by
henry2,
All angles are measured CW from +y-axis.
Given: M1 = 4,000kg, V1 = 12km/h[55o] E. of N.
M2 = 2500kg, V2 = 55km/h[270o].
V3 = 36km/h[270o] = Velocity of M1 after crash.
V4 = Velocity of M2 after crash.
Momentum before crash = Momentum after.
M1*V1+M2*V2 = M1*V3+M2*V4
4000*12[55o]+2500*55[270o] = 4000*36[270o]+2500V4
48,000[55o]+137,500[270o] = 144,000[270o]+2500V4
48,000[55o] - 6500[270o] = 2500V46
V4 = 19.2[55o] - 2.6[270o] = (19.2*sin55-2.6*sin270)+(19.2*cos55-2.6*cos270)i
V4 = 18.3+11.0i = 21.4km/h[59o].
Given: M1 = 4,000kg, V1 = 12km/h[55o] E. of N.
M2 = 2500kg, V2 = 55km/h[270o].
V3 = 36km/h[270o] = Velocity of M1 after crash.
V4 = Velocity of M2 after crash.
Momentum before crash = Momentum after.
M1*V1+M2*V2 = M1*V3+M2*V4
4000*12[55o]+2500*55[270o] = 4000*36[270o]+2500V4
48,000[55o]+137,500[270o] = 144,000[270o]+2500V4
48,000[55o] - 6500[270o] = 2500V46
V4 = 19.2[55o] - 2.6[270o] = (19.2*sin55-2.6*sin270)+(19.2*cos55-2.6*cos270)i
V4 = 18.3+11.0i = 21.4km/h[59o].
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