Answer:
The speed of the 3264 kg car before collision is 6.32 m/s
Explanation:
Let first car, mass m₁, moving east be moving with velocity, v₁ and second car, mass m₂, moving north be moving with velocity, v₂. Let the velocity of the two cars after collision be v₃
In a Collison (whether elastic or inelastic), momentum is always conserved.
Momentum before collision = momentum after collision.
Momentum before collision = m₁v₁ + m₂v₂
v₁ = (12.91î) m/s in vector form, m₁ = 2137 kg
v₂ = ?, v₂ = (v₂j) m/s in vector form, m₂ = 3264 kg
Momentum before collision = (2137)(12.91î) + (3264)(v₂j) = (27590î + 3264v₂j) kgm/s
Momentum after collision = (total mass of the cars after collision) × (velocity of the stuck-together cars after collision)
Total mass of the cars after collision = m₁ + m₂ = 2137 + 3264 = 5401 kg
Velocity after collision, v₃ = 6.38 m/s In the N36.8°E direction, put in vector form,
v₃ = (6.38 cos 36.8°î + 6.38 sin 36.8°j)
v₃ = (5.11î + 3.82j) m/s
Momentum after collision = 5401 (5.11î + 3.82j) = (27590î + 20641.4j) kgm/s
Momentum before collision = momentum after collision.
(27590î + 3264v₂j) = (27590î + 20641.4j)
Comparing components
3264v₂ = 20641.4j
v₂ = 6.32 m/s
Hope this Helps!!!
