Respuesta :
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
=[tex]\dfrac{1}{2}(mv^2 + Mv'^2)- \dfrac{1}{2}(mu^2 + Mu'^2)[/tex]
=[tex]\dfrac{1}{2}(1200 \times18^2 +9000 \times 20.93^2)- \dfrac{1}{2}(1200 \times 25^2 +9000\times 20^2)[/tex]
=9038 J
Answer:
Explanation:
Given
mass of car [tex]m=1200 kg[/tex]
initial speed of car [tex]v_1=25 m/s[/tex]
Final Speed of car [tex]v_2=18 m/s[/tex]
Mass of Truck [tex]M=9000 kg[/tex]
initial speed of Truck [tex]u_1=20 m/s[/tex]
Let [tex]u_2[/tex] be the speed of truck after collision
Conserving momentum
[tex]mv_1+Mu_1=mv_2+Mu_2[/tex]
[tex]1200\times 25+9000\times 20=1200\times 18+9000\times u_2[/tex]
[tex]30,000+1,80,00=21,600+9000\cdot u_2[/tex]
[tex]u_2=\frac{188400}{9000}[/tex]
[tex]u_2=20.93 m/s[/tex]
(b)Initial Kinetic Energy of car [tex]K.E._{1c}=\frac{1}{2}mv_1^2=3,75,000 J[/tex]
Initial Kinetic Energy of Truck [tex]K.E._{1T}=\frac{1}{2}Mu_1^2=1,800,000 J[/tex]
Final Kinetic Energy of car [tex]K.E._{2c}=\frac{1}{2}mv_2^2=1,94,400 J[/tex]
Final Kinetic Energy of Truck [tex]K.E._{2T}=\frac{1}{2}Mu_2^2=1,971,292.05 J[/tex]
Change in kinetic Energy=Initial Kinetic Energy-Final kinetic Energy
[tex]=(3,75,000+1,800,000)-(1,94,400+1,971,292.05)[/tex]
[tex]=9307.95 J[/tex]
