A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000

Question

A 1500-kg car traveling at 90 km/h toward the east suddenly collides with a 3000-kg car traveling at 60 km/h toward the south. The two cars stick together after the collision. What is the speed of the cars after collision?  

a) 14 m/s

b)22 m/s

c) 8.3 m/s

d) 17 m/s

A 900-kg car traveling 30.0° south of east at 12.0 m/s suddenly collides with a 750-kg car traveling north at 17.0 m/s. The cars stick together after colliding. What is the speed of the wreckage just after the collision?

a) 12.2 m/s

b) 17.3 m/s

c) 20.4 m/s

d) 25.0 m/s

e) 7.21 m/s

Pleas show work.

Explanation / Answer

1) a)14 m/s

vx = 1500*90/(1500+3000)

= 30 km/h

vy = 3000*60/(1500+3000)

= 40 m/s

v = sqrt(vx^2 + vy^2)

= sqrt(30^2 + 40^2)

= 50 km/h

= 50*5/18

= 14 m/s

2) a)12.2 m/s
Let East be +x axis.

let v is the speed of wreckage.

let

m1 = 900 kg
u1 = 12 m/s
u1x = 12*cos(30) = 10.39 m/s
u1y = -12*sin(30) =-6 m/s

m2 = 750 kg
u2x = 0
u2y = 17 m/s

Apply conservation of momentum in x direction

m1*u1x + m2*u2x = (m1+m2)*vx
900*10.39 + 0 = (900+750)*vx
==> vx = 900*10.39/(900+75)

= 9.59 m/s

Apply conservation of momentum in y direction

m1*u1y + m2*u2y = (m1+m2)*vy

900*(-6) + 750*17 = (900+750)*vy

==> vy =(900*(-6) + 750*17) /(900+75)

= 7.54 m/s

so, v = sqrt(vx^2 + vy^2)

= sqrt(9.59^2 + 7.54^2)

= 12.2 m/s <<<<<<------------Answer

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