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

Answer

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

           =$\dfrac{1}{2}(mv^2 + Mv'^2)- \dfrac{1}{2}(mu^2 + Mu'^2)$

           =$\dfrac{1}{2}(1200 \times18^2 +9000 \times 20.93^2)- \dfrac{1}{2}(1200 \times 25^2 +9000\times 20^2)$

           =9038 J

Answer 2

Answer:

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

$mv_1+Mu_1=mv_2+Mu_2$

$1200\times 25+9000\times 20=1200\times 18+9000\times u_2$

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

$u_2=\frac{188400}{9000}$

$u_2=20.93 m/s$

(b)Initial Kinetic Energy of car $K.E._{1c}=\frac{1}{2}mv_1^2=3,75,000 J$

Initial Kinetic Energy of Truck $K.E._{1T}=\frac{1}{2}Mu_1^2=1,800,000 J$

Final Kinetic Energy of car $K.E._{2c}=\frac{1}{2}mv_2^2=1,94,400 J$

Final Kinetic Energy of Truck $K.E._{2T}=\frac{1}{2}Mu_2^2=1,971,292.05 J$

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

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

$=9307.95 J$