Loaded Carts

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:

M v_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