## Conservation of Momentum

A loaded cart is coasting along a road at a constant speed v_{a}. Suddenly, part of its load (equal to 20% of its mass) is projected forwards, causing the cart to some to rest. If, instead, the same amount was projected backwards, it would cause the cart’s velocity to increase to a value v

_{b}. What is the ratio between v

_{b}and v

_{a}?

We will solve this problem, using conservation of momentum – in each of the two cases where the load is projected off the cart, the momentum of the cart – load system should remain unchanged.

If we take motion in the forward direction to be positive, the conservation of momentum for the first case can be written as:

- M v

_{a}= 0.8*M*0 + 0.2*M*v

_{c}

Where v_{c} is the velocity the projected load aquires.

Then, we can see that:

- v

_{c}= 5*v

_{a}

Now, going to the second case (where the mass is projected backwards), conservation of momentum gives us:

- M v

_{a}= 0.8*M*v

_{b}- 0.2*M*v

_{c}

The differences between this and the previous equation are the minus sign – this is to take into account that the velocity of the mass in this case goes in the other direction (the negative direction) – and the v_{b}, because in this case, the cart is not stationary after the ejection of the mass.

Now, putting the v_{c} we calculated earlier into this equation, we get:

_{a}= 0.8 M v

_{b}– 0.2 M 5 v

_{a}

2M v_{a} = 0.8 M v_{b}

v_{b}/v_{a} = 2/.8 = 1.25

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