Question

While driving down the street, your friend’s car was hit by a truck from behind. The mass of your friend’s car is 1500 kg while the truck is 1800kg. Your friend was driving 25 mph or 11.2 m/s. Assume that collision causes both cars to stick together and they do not supply additional power to the wheels after the collision. After the collision, the two cars slide to a stop. The coefficient of kinetic friction is 0.65 for the car and truck with the road.

(15 pts) If the cars slide a total of 15 m after the collision, what was the speed of the connected cars at the start of the slide immediately following the collision?

(15pts) How fast was the truck moving before the collision? If you do not know the speed of the connected cars after the collision from part a), assume you know it and solve algebraically.

Answer #1

let,

mass of the truck,m1=1500 kg

mass of the car, m2=1800 kg

speed of the car, u2=11.2 m/sec

coefficient of kinetic friction,uk=0.65

a)

after collision,

distance travelled, s=15m

and

force=uk*N

(m1+m2)*a==uk*(m1+m2)*g

==> a=-uk*g

a=-0.65*9.8

a=-6.37 m/sec^2

acceleration, a=-6.37 m/sec^2

and

v^2-u^2 =2*a*s

0-u^2=2*(-6.37)*15

===> u=13.834 m/sec

speed of the car and truck, just after collision, u=13.834 m/sec

b)

by using the conservation of momentum,

m1*u1+ m2*u2=(m1+m2)*u

1800*u1+1500*11.2=(1800+1500)*13.834

===> u1=16.029 m/sec

speed of the truck before the collision, u1=16.029 m/sec

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