A2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north. the cars stick together and move as a unit after the collision, at an angle of 48.0° north of east and at a speed of 5.98 m/s. find the speed of the 3,000-kg car before the collision.

Speed of the 3,000-kg car before the collision = 7.41 m/s

Explanation:

Here momentum is conserved.

A 2,000-kg car moving east at 10.0 m/s collides with a 3,000-kg car moving north.

Initial momentum = 2000 x  10 i + 3000 x v j = 20000 i + 3000 v j

The cars stick together and move as a unit after the collision, at an angle of 48.0° north of east and at a speed of 5.98 m/s.

Mass = 2000 + 3000 = 5000 kg

Velocity = 5.98 cos 48 i + 5.98 sin 48 j = 4 i + 4.44 j

Final momentum = 5000 x (4 i + 4.44 j) = 20000 i + 22220 j

We have momentum is conserved

       20000 i + 3000 v j = 20000 i + 22220 j  

       3000 v = 22220

        v = 7.41 m/s

 Speed of the 3,000-kg car before the collision = 7.41 m/s

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a=d/t

a="triangle"v/t

explanation:

a = acceleration