Respuesta :
Answer
given,
mass of two automobile = 1000 kg
speed of the two automobile = 20 m/s
let north be +y direction and east be +x directon.
a) from momentum conservation
m₁ v₁ + m₂ v₂ = (m₁ + m ₂) v
1000 × 20 - 1000 ×20 = (1000 + 1000)× v
v = 0 m/s [no direction]
b) from moment of momentum conservation in x-direction
1000 × 20 = (1000 + 1000)× v _x
v_x = 10 m/s
from moment of momentum conservation in x-direction
1000 × 20 = (1000 + 1000)× v _y
v_y = 10 m/s
Resultant = [tex]\sqrt{10^2+10^2}[/tex]
= 14.14 m/s
[tex]tan \theta = \dfrac{10}{10}[/tex]
[tex]tan \theta = 1[/tex]
[tex]\theta = 45^0[/tex]
The magnitude of the final speed is 14.14 m/s
And, the direction should be 45 degrees.
Calculation of the direction and speed:
Since
mass of two automobiles = 1000 kg
speed of the two automobiles = 20 m/s
let us assume north be +y direction and east be +x directon.
a) Now here we applied momentum conservation
m₁ v₁ + m₂ v₂ = (m₁ + m ₂) v
1000 × 20 - 1000 ×20 = (1000 + 1000)× v
v = 0 m/s [no direction]
b) Now from moment of momentum conservation in the x-direction
1000 × 20 = (1000 + 1000)× v _x
v_x = 10 m/s
Now from moment of momentum conservation in x-direction
1000 × 20 = (1000 + 1000)× v _y
v_y = 10 m/s
[tex]Resultant = \sqrt{10^2 + 10^2}\\\\= 14.14 m/s[/tex]
Now
[tex]tan \theta = 10 \div 10\\\\tan \theta = 1 \\\\tan \theta = 45^{\circ}[/tex]
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