Perfectly Inelastic Collisions?

Question

A dry cleaner throws a 22kg bag of laundry onto a stationary 9kg cart. The cart and the laundry bag begin moving at 3 m/s to the right. Find the velocity of the laundry bag before the collision.

Answer 1

You conserve the momentum.... 22 * v = (22+9) * 3 and v = 4.22 m/s (to the right)

Answer 2

The Law of conservation of momentum is at work here.
The initial momentum of the system (the bag of laundry and the cart) must equal the final momentum of the system.

Since the cart is initially at rest, the only component of the system which posses a non-zero momentum is the bag of laundry. Therefore, the total initial momentum of the system is the momentum of the bag.
Initial Momentum = mass of bag * velocity of bag
P_initial = m_bag * v_bag_initial

After the collision, both the bag of laundry and the cart are moving and therefore both possess momentum. The final momentum of the system is the sum of the momentums of each component (the momentum of the bag plus the momentum of the cart).
Final momentum = mass of bag * velocity of bag + mass of cart + velocity of cart
P_f = m_bag * v_bag_f + m_cart * v_cart_f

We also know that the final momentum will equal the initial momentum so we can set the two above equations equal,
P_i = P_f
m_bag * v_bag_initial = m_bag * v_bag_f + m_cart * v_cart_f

We are given some values which we can plug into this equation.
We know the mass of the bag is 22 kg and the mass of the cart is 9 kg. We know the final momentum of the bag-cart combination is 3 m/s.
Plugging these values in we get,

(22 kg) * V = (22 kg) * (3 m/s) + (9 kg) * (3 m/s)
Where V is now the initial velocity of the bag which we do not know.
Solving for V,
V = ((22 kg) * (3 m/s) + (9 kg) * (3 m/s)) / (22 kg)

V = (93 kg m/s) / (22 kg)
V = 4.23 m/s.

Thus, the bag was initially traveling at a speed of 4.23 m/s prior to the collision with the cart.

Answer 3

I'm trying to be creative here. Stifling, stifling to be sure.