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A 1.80 -kg falcon catches a 0.650 -kg dove from behind in midair. What is their velocity after impact if the falcon's velocity is initially 28.0 $\mathrm{m} / \mathrm{s}$ and the dove's velocity is 7.00 $\mathrm{m} / \mathrm{s}$ in the same direction?

$v ^ { \prime } = 22.43 \mathrm { m } \mathrm { s } ^ { - 1 }$

03:13

Shital R.

Physics 101 Mechanics

Chapter 8

Linear Momentum and Collisions

Moment, Impulse, and Collisions

Hope College

University of Sheffield

McMaster University

Lectures

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here, we're going to do an example of conservation of momentum. So a reminder that momentum is conserved if there are no external forces acting on a system. Um an alternative way of saying this is that if there are external forces, they are negligible compared to the interactions between the objects and the system. Um and typically you have conservation of momentum with interacting objects which interact fear Newton's 3rd law pairs. That is two things collide and the force that one of object exerts on another is equal and opposite to the others force exerted on it. So, a very typical situation is with collisions, even if there was friction in the road when two cars collide, the collision forces are so large compared to that friction that you can basically neglect the friction. So here's our interaction. We have two birds. One is a big hawk Flying very quickly. Let's see. We'll call him mass. one 1.8 kg flying at a velocity of 28 m/s. Very fast for a bird. And he catches up to a little tiny dove, okay. Which will call mass to um that looks more like a bullet but it's a tiny dove. Um That's a mess of .650 kg. It is trying desperately to escape With seven m of a second velocity. Okay, so we are going to conserve momentum. What is momentum, momentum is a vector product of mass times velocity. We have things just happening in the X direction. So we are going to be interested just in X components of velocity. Now, what does it mean to conserve momentum? It means that the some of the P. X. Is in the system initially before the collision equals the sum of them. Finally. So here we have two birds and we want to add both of their masses. I am on V one times there, they're X components of velocity which are both positive. And at the end we can show a final Okay, so that's initial. But the final situation is There's just one bird with the dove and its beak. Okay, so they're off together flying with final speed and the masses are together. Now that's the final situation, by the way, this is something known as an in elastic collision which means that the objects stick together and go off as a solid single object. It's called an any elastic collision because nobody bounced off of someone else. If the two birds smacked into each other and bounced off of each other, that could be partially elastic. But anyway, we we've got all our quantities, so let's just put those in. Uh huh. Yeah and those half units of kilogram beat her per second on the other side. We add up our kilograms and it's the v final that we are trying to find. So we will solve that. And I essentially get the final is 55 Divided by 2.45 meters per second, Simplifying 22.4 meters per second, which does make some sense that the hawk had to um slow down to be able to carry off, uh, or mass with it.

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