Question

Two-car, 2-D collision. Two cars with masses m1 = 1200 kg and m2 = 1500 kg are approaching an intersection. Car 1 has a velocity v1 = 15.0 m/s towards the east and car 2 has a velocity v2 = 20.0 m/s towards the north. The two cars reach the intersection at the same time and collide and lock bumpers and after the collision travel as a single wreckage.

(a) What is the magnitude and direction of their velocity after the collision?

(b) What is the impulse delivered by car 2 to car 1? Give the impulse in vector notation.

Answer 1

Given

car with a mass m₁ = 1200 kg

initial speed of v₁= 15 m/s

Another car with mass m₂ = 1500kg

initial speed of v₂= 20 m/s

Apply the law of linear momentum is along x-axis

m₁v₁ =(m₁+ m₂) v_f cosθ

Apply the law of conservation of momentum along y -axis

m₂ v₂ = ( m₁+m₂)  v_f sinθ

tanθ = m2v2/m1v1

θ = 59.036 with x-axis

m₁v₁ =(m₁+ m₂)v_f cosθ

(1200)(15) = (1200+1500 )V_f cos59.036^o

V_f = 12.957 m/s

therefore,the speed of the wreckage is 12.957 m/s

direction of the wreckage after the collision is 59.036° above the x-axis

impulse delivered = change in momentum = 1200(15-12.957)=2451.6 Ns

In case of any doubt, please do comment sir.