A 75 kg archer on ice skates is standing at rest on very smooth ice. He shoots a 450 g arrow horizontally. When released, the arrow reaches a speed of 110 m/s in 0.25 s. Assume that the force of the bow string on the arrow is constant. b. What is the archer's recoil speed?

Video transcript

Hey, everyone. So this problem, we have a researcher that has a mass of 82 kg standing on a frictionless wet ground holding a gun. The researcher fires 25 g bullet horizontally. The bullet acquires a speed of 650 m per second. In 1.2 seconds were asked to determine the rebound velocity of the researcher assuming a constant force on the bullet. So to approach this problem, we're gonna draw two different free body diagrams, one for the bullet and one for the researcher. So the F B D for our bullet, it's just going to look like this. We have one force acting in the positive horizontal direction you wanna call that F sub G or the force of the gun. Our free body diagram for our researcher is going to look like this. We have one force in the negative horizontal direction and we don't have a friction force because it's a frictionless wet ground. So the only force acting on the researcher is the force from the gun. Now, we can recall based on Newton's third law of equal and opposite reactions that the force of the gun on the bullet is going to be equal and opposite of the force of the bullet on the gun. So this F sub G that we can find using our first free body diagram, the magnitude is going to be the same for the force that's acting on the researcher. So back up to our bullet, we can recall from Newton's second law that F equals M A where mass for the bullet was given to us as 25 g. And the problem, so I'm gonna rewrite that 0.25 kg to keep us in standard units. And while we don't have acceleration from the problem, we do have velocity in time. So we can recall that acceleration is given by delta V over delta T. So we can plug that in to our force equation here. And this is F sub G. So we've got FFG equals 0. kg times delta V over delta T. That's gonna be 650 m per second over 1.2 seconds, plug that in to our calculator And get 13.5 newtons. So now when we take the, some of our force is equal to mass times acceleration. In our second free body diagram will have F sub G equals um A and this SMG is gonna be negative acting and the negative horizontal or negative X direction. And in this case, the mass of our researcher was given to us and the problem as 82 kilograms. So we will use this to solve for acceleration. So a equals negative 13.5 Newtons divided by 82 kilograms, Which equals negative 0. 65 m/s. And again, we're calling that A is equal to delta V over delta T. Now we're trying to solve for the velocity so we can rearrange that equation to be the equals A times T where we have T from the problem. And we've just solved for our acceleration. So negative 0.165 m per second squared times 1.2 seconds Equals -0.19, 8 meters per second. So that is our final answer for this problem. And if we look at our choices that aligns with answer C so answer C is the correct choice for this problem. And that's all we have for this one. We'll see you in the next video.