Question

A 1200-kg car traveling initially with a speed of 25.0 m/s in an easterly direction crashes into the rear end of a 9,000-kg truck moving in the same direction at 20.0 m/s. The velocity of the car right after the collision is 18.0 m/s to the east.a. What is the velocity of the truck right after the collision?b. How much mechanical energy is lost in the collision? Account for this loss in energy.

Answer 1

Step-by-step explanation:

Given

mass of car
`m=1200 kg`

initial speed of car
`v_1=25 m/s`

Final Speed of car
`v_2=18 m/s`

Mass of Truck
`M=9000 kg`

initial speed of Truck
`u_1=20 m/s`

Let
`u_2` be the speed of truck after collision

Conserving momentum

`mv_1+Mu_1=mv_2+Mu_2`

`1200* 25+9000* 20=1200* 18+9000* u_2`

`30,000+1,80,00=21,600+9000\cdot u_2`

`u_2=(188400)/(9000)`

`u_2=20.93 m/s`

(b)Initial Kinetic Energy of car
`K.E._(1c)=(1)/(2)mv_1^2=3,75,000 J`

Initial Kinetic Energy of Truck
`K.E._(1T)=(1)/(2)Mu_1^2=1,800,000 J`

Final Kinetic Energy of car
`K.E._(2c)=(1)/(2)mv_2^2=1,94,400 J`

Final Kinetic Energy of Truck
`K.E._(2T)=(1)/(2)Mu_2^2=1,971,292.05 J`

Change in kinetic Energy=Initial Kinetic Energy-Final kinetic Energy

`=(3,75,000+1,800,000)-(1,94,400+1,971,292.05)`

`=9307.95 J`

Answer 2

given,

mass of car (m) = 1200 Kg

speed of cur, u = 25 m/s

mass of truck(M) = 9000 Kg

speed of truck, u' = 20 m/s

v = 18 m/s

a) using conservation of momentum

m u + M u' = m v + M v'

1200 x 25 + 9000 x 20 = 1200 x 18 + 9000 x v'

9000 v' = 188400

v' = 20.93 m/s

b) To calculate losses, we will find the kinetic energies before & after collision. Any difference would give us the losses (in energy form).

(K E)₁= (K E)₂+ Losses

losses = (K E)₂ - (K E)₁

=
`(1)/(2)(mv^2 + Mv'^2)- (1)/(2)(mu^2 + Mu'^2)`

=
`(1)/(2)(1200 *18^2 +9000 * 20.93^2)- (1)/(2)(1200 * 25^2 +9000* 20^2)`

=9038 J