Question

A 8500 kg truck moving at 11 m/s collides with a 1500 kg car at rest. The coefficient of friction is .53.

a) How fast is the truck going immediatley after hitting the car

b) How far do the truck and car travel before they stop. (Use Kinematics)

c) Using the conservation of energy equation, find for the same distance.

Answer 1

here,

mass of truck , M = 8500 kg

mass of car , m = 1500 kg

initial speed of truck , v0 = 11 m/s

let the final speed of the combination be v

using conservation of momentum

M * v0 = (m + M) * v

8500 * 11 = ( 8500 + 1500 ) * v

v = 9.35 m/s

a)

speed of truck after hitting the car is 9.35 m/s

b)

accelration due to friction , a = - u * g

a = - 5.194 m/s^2

let the distance taken to stop be d

using third equation of motion

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

0 - 9.35^2 = - 2 * 5.194 *

d = 8.42 m

c)

using work energy theorm

work done by frictional force = kinetic energy lost

u * (m+M) * g * d = 0.5 * (m + M) * v^2 - 0

0.53 * 9.8 * d = 0.5 * 9.35^2

d = 8.42 m