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Physics 20 Online
OpenStudy (batgirl13):

PLEASE HELP!!!! ASAP!!!! 17. If a stationary cart is struck head-on by a cart with twice mass of the stationary one and a velocity of 5 m/s, what will be the new velocity of the stationary cart if the collision is inelastic? Show all calculations leading to an answer.

OpenStudy (farcher):

This problem only has a solution if the assumption that inelastic means the two carts stick together and then move off together. If that assumption is not made then you have one equation (the conservation of momentum) and two unknowns (the final velocity of each of the two carts).

OpenStudy (batgirl13):

So if we make that assumption about the problem how would I solve and explain it? @Farcher

OpenStudy (farcher):

You use conservation of momentum assuming that no external forces are acting. You assume that after collision the carts stick together. You know the initial momentum.. You know the total mass after collision. You put initial momentum equal to final momentum and work out the final speed. In words state the direction in which the two carts will be moving after the collision relative to the initial direction of motion of the moving cart.

OpenStudy (batgirl13):

What is the total mass after the collision. I'm having a really hard time with this problem, sorry...

OpenStudy (anonymous):

You actually don't know the final mass. All you know is the moving cart is twice the mass of the stationary cart. The mass after the collision would be the total mass of the carts, or three times the mass of the stationary cart. You could use X to denote the mass of the stationary cart, 2X for the mass of the moving cart and 3X for the combined mass after the collision. You'll find that X will eventually cancel out of the equation. If you are not comfortable with using a variable for the masses, you could pick a value. Use 1kg for the stationary cart, 2kg for the moving cart, and 3kg for the mass after the collision. Or use 10kg, 20kg and 30kg. You'll find that actual masses don't matter, just the ratio of masses.

OpenStudy (batgirl13):

okay I understand! thank you thank you!!

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