Conservation of momentum: vertical shot

AI Thread Summary
A bullet weighing 0.028 kg is fired vertically at 165 m/s into a stationary 0.15 kg baseball, which it embeds into upon collision. Using conservation of momentum, the final velocity of the combined mass is calculated to be approximately 25.96 m/s. The height reached by the bullet and baseball together is determined using the equation for vertical motion, resulting in a corrected height of 34.4 meters after resolving an initial calculation error. The discussion highlights the importance of careful arithmetic in physics problems. The final answer confirms the correct height achieved after the collision.
mandy9008
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Homework Statement


A 0.028 kg bullet is fired vertically at 165 m/s into a 0.15 kg baseball that is initially at rest. How high does the combined bullet and baseball rise after the collision, assuming the bullet embeds itself in the ball?


Homework Equations


mv + mv = mv + mv


The Attempt at a Solution


m1 = 0.028 kg
v1 = 165 m/s
m2 = 0.15 kg
v2i = 0 m/s

(0.028 kg)(165 m/s) + (0.15 kg)(0 m/s) = (0.178 kg) vf
vf=25.96 m/s

v2=vo2 - 2gy
25.962 = - 2(9.8 m/s2)y
y=693.5m
 
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mandy9008 said:
v2=vo2 - 2gy
25.962 = - 2(9.8 m/s2)y
y=693.5m

That does not give 693.5 m. Recheck it.
 
ah, i subtracted instead of divided. man, i feel silly now! thank you! when I divided, i got 34.4m, which is correct :)
 
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