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A gun has a recoil speed of 2 m/s when firing. If the gun has a mass of 2kg and the bullet has a mass of 10g (0.01 kg) what speed does the bullet come out at?

The gun has zero total momentum before firing and afterwards the gun has negative acceleration.

So far:

Conservation of momentum: $m_1v_1 = m_2v_2.$

We have the recoil speed of $2\,$m/s. The mass of the gun is equal to $2\,$kg.

Plus the bullet's total mass which is 0.01 kg.

$$2 \frac{m}{s}\cdot 2.01\,\text{kg} + (0.01\,\text{kg} \cdot v)$$ $$=4.02\,\text{kg}\frac{m}{s} + (0.01\,\text{kg} \cdot v)$$

That's as far as I can go.

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closed as off-topic by Brandon Enright, ACuriousMind, Neuneck, John Rennie, Kyle Kanos Feb 20 '15 at 13:40

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up vote 0 down vote accepted

Few remarks

Consider the whole system, this is , gun and bullet. This is an isolated system, so the net momentum is constant. In particular before firing the gun, the net momentum is zero. The conservation of momentum reads

$$0=m_{1}v_{1}+m_{2}v_{2} $$

If the call $1$ to the gun and $2$ the bullet:

$$\boxed{v_{2}=-\displaystyle\frac{m_{1}v_{1}}{m_{2}}} $$

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Just before I saw this I did this equation. But, thanks. This has compounded my answer. – lrobb Mar 17 '13 at 17:11

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