A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448-kg hay wagon moving with velocity v2=13.8i^+10.2j^m/s. If the two stay together, what is their velocity?
Express your answer in meters per second in terms of i^ and j^. Use the 'unit vector' button to denote unit vectors in your answer.

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 i + 12.7 j ) m/s) + (448 kg) ((13.8 i + 10.2 j ) m/s) = ((1250 + 448) kg) v

where v is the velocity of the system. Solve for v :

v = ((1250 kg) ((36.2 i + 12.7 j ) m/s) + (448 kg) ((13.8 i + 10.2 j ) m/s)) / (1698 kg)

v ≈ (30.3 i + 12.0 j ) m/s

the answer is b

explanation:

no sign of dia of fuel cell