Due to momentum conservation during the collision, we know that the final moment

Question

Due to momentum conservation during the collision, we know that the final momentum of the system will also be 1.5 times 10^4 kg m/s. This principle allows us to calculate what will happen after the collision, and predict the speed and direction of the objects after they collide. In this lab, we will test momentum conservation by performing 10 collisions of loaded or unloaded carts on a nearly frictionless track. If a football player, mass of 80 kg, is running to the right at a speed of 10 m/s, what momentum? Consider a system of two carts. One cart, mass of 0.500 kg, is at rest. The other cart has the same mass, and is moving at 18 m/s to the left, towards the first cart. What is the total momentum of the system, before the collision? Read through this lab and the worksheet with the 10 trials. In the analysis section of the lab, you will be asked to graph the initial momentum values against the final momentum values. You will have 10 data points. If a straight line were fit to these data points, what do you expect the slope of this line to be, if momentum is conserved during the collisions?

Explanation / Answer

given,

mass = 80 kg

speed = 10 m/s

momentum = mass * speed

momentum = 80 * 10

momentum of footballer = 800 kg.m/s

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