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

the independent variable, the one that is intentionally manipulated, is whether or not the car hits the barrier.

the dependent variable, the one that you measure the response in, is the distance that the washers travel.