What is the common velocity of the arrow

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The problem involves an arrow of mass 100 g and an apple of mass 80 g, with the arrow initially traveling at 40 m/s. Upon impact, the arrow and apple stick together, requiring the calculation of their common velocity post-collision using the conservation of momentum principle. The initial momentum of the arrow is calculated as 4 kg·m/s, and the momentum after impact must equal this value. The correct equation for the combined mass after the collision is (0.1 + 0.08)v = 4, leading to a common velocity of approximately 22 m/s. The confusion arises from misapplying the momentum conservation equation, emphasizing the importance of correctly accounting for the combined mass.
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Homework Statement



An arrow of mass 100 g is fired with an initial horizontal velocity of 40 m s–1 to the right at an apple of mass 80 g that is initially at rest on a horizontal surface. When the arrow strikes the apple, the two objects stick together. What is the common velocity of the arrow and apple after the impact?

Homework Equations


Momentum = mass * velocity


The Attempt at a Solution



Initial Momentum of the Arrow
p=m*v
=(.1)(40)
= 4

4=.08* v
therefore v= 50 m/s

the answer says it's 22m/s
 
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4=.08* v
therefore v= 50 m/s
This is wrong.
When the arrow strikes the apple, the two objects stick together.
So rewrite the above equation.
 


I don't get it
If the momentum is 0 then the velocity is 0
 


Momentum after impact = ( 0.1 + 0.08)*v
 


m1v1 = (m1 + m2)v
 
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