Calculating Velocity After Inelastic Collision

[ally1h]

Homework Statement

A 2.0 kg object is moving at 5.0 m/s NORTHWEST. It strikes a 6.0 kg object that is moving SOUTHWEST at 2.0 m/s. The objects have a completely inelastic collision. The velocity of the 6.0 kg object post collision is:

Homework Equations

m1v1+m2v1 = m1v2+m2v2

The Attempt at a Solution

I know the answer to be 2.33 m/s at 14 degrees North of West. I don't know how to get there. I THOUGHT, since the collision is inelastic, the equation is m1v1+m2v1=(m1+m2)v, and solve for v. That tells me the velocity of both the objects to be 3.2 m/s. Then I think I have to approach this collision from a 2 dimension perspective... but without angle to go off of, I am lost! Please help!

 

[Doc Al]

You most definitely must treat momentum as a vector. Treat north-south and east-west components separately. You are given all the angles that you need--translate southwest and northwest into angles.

 

[baba89]

I would first confirm that the given information is correct and that the objects are indeed moving in the directions stated. Next, I would use the conservation of momentum equation, m1v1 + m2v1 = (m1 + m2)v, to solve for the final velocity, v.

Plugging in the given values, we get (2.0 kg)(5.0 m/s) + (6.0 kg)(-2.0 m/s) = (2.0 kg + 6.0 kg)v.

Simplifying, we get 10.0 kg*m/s = 8.0 kg*v.

Dividing both sides by 8.0 kg, we get v = 1.25 m/s. This is the final velocity of the combined objects after the collision.

To determine the direction of this velocity, we can use the concept of vector addition. Since the objects are moving in opposite directions, the final velocity will be somewhere between the two initial velocities.

To find the angle, we can use the following equation: tan θ = (m1v1 sin θ1 + m2v2 sin θ2) / (m1v1 cos θ1 + m2v2 cos θ2).

Plugging in the values, we get tan θ = [(2.0 kg)(5.0 m/s)(sin 45°) + (6.0 kg)(2.0 m/s)(sin 225°)] / [(2.0 kg)(5.0 m/s)(cos 45°) + (6.0 kg)(2.0 m/s)(cos 225°)].

Simplifying, we get tan θ = [5.0 kg*m/s + (-6.0 kg*m/s)] / [5.0 kg*m/s + (-6.0 kg*m/s)].

This gives us tan θ = 1, which means θ = 45°.

Therefore, the final velocity of the 6.0 kg object is 1.25 m/s at 45° North of West. This confirms the given answer of 2.33 m/s at 14° North of West.

In conclusion, by using the conservation of momentum equation and vector addition, we were able to calculate the final velocity of the 6.0 kg object after the inelastic