Cofused with this Momentum problem

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In the discussion about a momentum problem in a combustion engine, a participant calculates the speed of gas molecules after an explosion, initially arriving at 6,000 m/s but noting the correct answer is 3,000 m/s. The confusion arises from the application of the impulse-momentum formula, where the participant uses ft = mv but misinterprets the change in momentum. The key point is that the change in momentum should account for the fact that the gas molecules bounce off the piston, effectively doubling the final momentum in the calculation. Clarification is sought on how to correctly apply the concept of momentum change to arrive at the correct speed of 3,000 m/s. Understanding the relationship between the momentum of the gas and the piston is crucial for solving the problem accurately.
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19. Incorrect In a combustion engine, gas is burned. The resulting explosion produces a force that drives the pistons in the engine. The force of the explosion on the piston is due to the change in momentum of the gas molecules. A 0.4-g sample of gas produces a force of 2400 N in an explosion that lasts 10e-3 seconds. What must be the speed of the gas molecules?

My answer: 6 x 10e3 m/s
The correct answer: 3 x 10e3 m/s

I used the impluse force = change in momentum forumla

ft = mv

2400(10e-3) = .0004(x)
and solved to get 6e3 m/s

What am I missing?
 
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the molecules don't stick to the piston, they bounce off.
so the change of momentum is ( p_f - p_i ) double the final momentum.
 
Err...

I get that...but how is the answer 3e3 m/s ?

could you put some numbers into the explanation, because conceptually I'm not getting why the change the velocity is 3e3??
 
suppose the molecule is above the piston, calling upward as positive.
v_i is negative "v" , but v_f is positive "v".
The change of momentum is m times 2"v".

The change of momentum of the molecules is the negative of
the change of momentum of the piston (I expect you DID get this).
 

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