A 1000-kg car is traveling north at 20.0 m/s. A 1500-kg car is traveling north at 36.0 m/s. The 1500-kg collides with the rear of the 1000 - kg car and they lock together. Ignoring external forces acting during the collision, what is the velocity of the cars immediately after the collision? (a) 56.0 m/s north (b) 29.6 m/s north (c) 13.6 m/s south (d) 8.00 m/s south (e) 28.0 m/s north

Question

Wilson

Physics

15 March, 06:25

A 1000-kg car is traveling north at 20.0 m/s. A 1500-kg car is traveling north at 36.0 m/s. The 1500-kg collides with the rear of the 1000 - kg car and they lock together. Ignoring external forces acting during the collision, what is the velocity of the cars immediately after the collision? (a) 56.0 m/s north (b) 29.6 m/s north (c) 13.6 m/s south (d) 8.00 m/s south (e) 28.0 m/s north

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Belinda Townsend · Answer 1

B

Explanation:

Use conservation of momentum: Total momentum of the system before is the same as the total momentum after the collision. Since both are moving north, the momentum of both cars is in the same direction, and the total momentum before collision is 1000kg*20m/s + 1500kg*36m/s = 74000kg-m/s.

This is the same amount of momentum after collision, and since they are locked together, their mass is added. The velocity can be found by: 74000kg-m/s : 2500kg = 29.6m/s.

momentum = mass*velocity