Head-on Collision: 4.5 kg & 11 kg Objects

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In a head-on collision between a 4.5 kg object moving at 4.0 m/s and an 11 kg object moving at 3.0 m/s, the 11 kg object comes to a complete stop. To find the final speed of the 4.5 kg object, the conservation of momentum must account for the direction of motion, resulting in the equation: (4.5 kg * 4.0 m/s) - (11 kg * 3.0 m/s) = (4.5 kg * Vf). The initial momentum should be calculated with one object's momentum as negative due to its opposite direction. Additionally, to determine if the collision is elastic, the conservation of energy should be checked.
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A 4.5 kg object with a speed of 4.0 m/s collides head-on with a 11 kg object moving toward it with a speed of 3.0 m/s. The 11 kg object stops dead after the collision.

(a) What is the final speed of the 4.5 kg object?
(b) Is the collision elastic?

so i tried initial momentum = final momentum but I get the wrong answer.
(4.5 kg)(4.0 m/s) + (11 kg)(3.0 m/s)=(4.5 kg)(Vf)
and solving for Vf = 11.333 m/s, but that is the wrong answer.
 
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Take into account that they have different directions. One momentum will be positive while the other negative.
 
Conservation of Momentum is the correct property to use, but you forgot to factor in that the two objects are moving at different directions before the collision, so initial momentum = 4.5kg * 4 m/s - 11kg * 3 m/s

For the second part, you should just see if Conservation of Energy is also upheld in the collision.

~Lyuokdea
 
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