Question

A 2200 kg car moving east at 10.0 m/s collides with a 3000 kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 36.0° north of east and at a speed of 5.23 m/s.
a) Find the speed of the 3000 kg car before the collision.
b) What is the decrease in kinetic energy during the colision?

Answer 1

Answer

given,

mass of car 1, m₁ = 2200 Kg

Speed of car in east, v₁ = 10 m/s

Mass of the car 2, m₂ = 3000 Kg

Speed of car 2 in north, v₂ = ?

Speed of the system after collision, v = 5.23 m/s

Angle of velocity after. θ = 36.0°

Velocity Horizontal component

v_x = v cos θ

v_x = 5.23 cos 36°

v_x = 4.23 m/s

Vertical velocity component

v_y = v sin θ

v_y = 5.23 sin 36°

v_y = 3.074 m/s

a) writing conservation of equation in north direction

m₂v₂ = (m₁+m₂)v_y

3000 x v₂ = (3000+2200) x 3.074

v₂ = 5.33 m/s

b) Initial Kinetic energy

 $KE_1 = \dfrac{1}{2}m_1v_1^2 + \dfrac{1}{2}m_2v_2^2$

 $KE_1 = \dfrac{1}{2}\times 2200\times 10^2 + \dfrac{1}{2}\times 3000\times 5.33^2$

$KE_1 = 152613.35\ J$

Final Kinetic energy

  $KE_2 = \dfrac{1}{2}(m_1+m_2)v^2$

  $KE_2 = \dfrac{1}{2}\times 5200\times 5.23^2$

  $KE_2 =71117.54\ J$

Hence, Decrease in Kinetic energy

$\Delta KE = KE_1-KE_2$

$\Delta KE = 152613.35-71117.54$

$\Delta KE = 81495.81\ J$

Decrease in Kinetic energy after collision is equal to $\Delta KE = 81495.81\ J$