# Conservation of Momentum from Recoil Speed

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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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

solution $m_1v_1 + m_2v_2$ = $m_1v_a + m_2v_b$ since they both are at rest intially so LHS turns to zero and the expression becomes $m_1v_a + m_2v_b = 0$ now since the recoil is in opposite direction to the velocity of the bullet so let us take direction of the velocity of the bullet as positive so the direction of recoil will become negative and the expression will become$m_1v_a -m_2v_b = 0$ so $m_1v_a = m_2v_b$ now substitute the values and you will get the answer which I found was $400m/s$

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