Inelastic collision: final velocity after collision

AI Thread Summary
The discussion focuses on calculating the change in velocity of a 1000-kg car after a 9.0-g bug collides with it. The initial velocities are 19 m/s for the car and -1.5 m/s for the bug. Participants suggest using the conservation of momentum equation to find the final velocity, leading to a corrected change in velocity calculation. The final answer arrived at is approximately -1.845 x 10^-4 m/s, emphasizing the importance of significant figures in reporting results. The conversation highlights common pitfalls in calculations and the need for careful simplification to avoid errors.
emily081715
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Homework Statement


You are driving your 1000-kg car at a velocity of(19 m/s )ι^ when a 9.0-g bug splatters on your windshield. Before the collision, the bug was traveling at a velocity of (-1.5 m/s )ι^.
What is the change in velocity of the car due to its encounter with the bug?

Homework Equations


pi = pf
m1v1 + m2v2 = (m1 + m2)v

The Attempt at a Solution


i attempted to solve for the change in velocity using the equation outlined below and got answer of -3.0x10^-5. i know this answer is inncorrect but i cannot seem to locate my source of error
Δv=(1.9x10^4)+ (-1.35x10^-2) -19 m/s
1000 +0.009
 
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emily081715 said:

Homework Statement


You are driving your 1000-kg car at a velocity of(19 m/s )ι^ when a 9.0-g bug splatters on your windshield. Before the collision, the bug was traveling at a velocity of (-1.5 m/s )ι^.
What is the change in velocity of the car due to its encounter with the bug?

Homework Equations


pi = pf
m1v1 + m2v2 = (m1 + m2)v

The Attempt at a Solution


i attempted to solve for the change in velocity using the equation outlined below and got answer of -3.0x10^-5. i know this answer is inncorrect but i cannot seem to locate my source of error
Δv=(1.9x10^4)+ (-1.35x10^-2) -19 m/s
1000 +0.009
Bring the terms of the difference to common denominator, then 1.9x10^4 cancels, and you avoid rounding errors.
 
ehild said:
Bring the terms of the difference to common denominator, then 1.9x10^4 cancels, and you avoid rounding errors.
i am unsure what you mean for me to do/ how to do that
 
You calculate the difference ##\frac {19\cdot 1000- 1.5\cdot 0.009}{1000+0.009}-19 = \frac{19\cdot 1000- 1.5\cdot 0.009-19(1000+0.009)}{1000+0.009}##
Expand and simplify.
 
ehild said:
You calculate the difference ##\frac {19\cdot 1000- 1.5\cdot 0.009}{1000+0.009}-19 = \frac{19\cdot 1000- 1.5\cdot 0.009-19(1000+0.009)}{1000+0.009}##
Expand and simplify.
i got an answer of -1.8449 x10^-4
 
emily081715 said:
i got an answer of -1.8449 x10^-4
It looks correct, but do not keep more than 3 significant digits.
 
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