A 20 kg wood ball hangs from a 2.0-m-long wire. The maximum te...

Question

A 20 kg wood ball hangs from a 2.0-m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A 1.0 kg projectile traveling horizontally hits and embeds itself in the wood ball. What is the greatest speed this projectile can have without causing the wire to break?

Accepted Answer

Hey, everyone in this problem, an 18 g dart tip is shot horizontally from a toy gun and collides with a 360 g polyester sphere that's suspended by a 0.75 m light string. Can it gets stuck to it? We're told that the breaking strength of the string is five newtons. And we're asked to find the tips, maximum speed just before the collision that will not provoke a string crack. We're given four answer choices. Option A 11.5 m per second. Option B 21 m per second. Option C 33.6 m per second and option D 42.5 m per second. So let's start by just drawing a little diagram. So we have this sphere hanging from a string vertically and our dirt tip is going to be shot horizontally into that sphere after it hits that sphere, it's gonna cause it to swing on that string with the dirt tip stuck inside the sphere. All right. So we wanna find the maximum speed that will not provoke a string crack. So the first thing we're gonna do is actually find the speed of this combination after the collision. So we have this collision problem, we want to think about the conservation of momentum. So we know that our initial momentum P is gonna be equal to our final momentum P F. Now we have two things in our system, we have the dart tip and we have this sphere. OK. So we have two components that are gonna make up our momentum. So for each of these, we have to consider both. So we have P not D, the initial momentum of the dart plus peanut as the initial momentum of the sphere is equal to P F D plus P F S. OK. So we're using S for sphere D for dart. Now we're called that momentum is equal to mass multiplied by velocity. So for each of these terms, we have the corresponding mass multiplied by the corresponding velocity M DVD plus MS VA S is equal to. Now, in the final case, OK, the dart is going to stick into the sphere. So they're gonna be moving together. So we're gonna have one combined mass MD plus MS and they're gonna be moving together at the same speed, the same velocity. So we have that combined mass multiplied by the final velocity V F which is that velocity of the dart tip and sphere stuck together. Now, initially, the sphere is suspended by this string. And so it's not gonna be moving, its initial velocity is going to be zero. And So the second term on the left hand side goes to zero filling in everything else. We know, we know that the mass of the dart is 18 g. Yeah. So we have 18 g multiplied by the initial velocity of the dirt is going to be equal to the mass of the dart. 18 g plus the mass of the sphere. 360 g multiplied by the final velocity V F we want to solve for that final velocity in terms of the initial velocity. OK. We're trying to figure out what initial velocity we can have. And so let's write that in terms of that value. So we get that the final velocity V F is equal to g multiplied by V not D divided by 378 g. Now, until this point, you might have been thinking, why are you using grams? We always convert into our standard unit of kilograms and you do want to do that most of the time. In this case, we end up with grams divided by grams. So that unit of gram is gonna divide out. We don't need to worry about the units. OK? So all we needed was for both of them to be the same, we could have converted them both into kilograms and we would have gotten the same ratio at the end that unit of kilogram would have canceled out, ok? So that's why we did it. In this case. But you do have to be careful about your units because that's not always the case. Now, if we simplify this, OK, we have 18 divided by 378 both of those numbers are divisible by 18. So if we divide the numerator and denominator by 18, we get that this is equal to one divided by multiplied by V not D OK. All right. So we've written the final velocity in terms of the initial velocity of the dark. Now, we need to find the maximum speed that will not provoke a provoke a string crack. So let's look into when we're going to get a string crack. OK? We have this breaking strength. Let's draw a free body diagram and think about the forces acting. You draw everybody diagram. OK. We have this upward force of tension and we have this downward force of gravity. Now, this tension force is pointing up towards the center gravity is pointing downwards. OK? And this free body dia diagram is immediately as that dart hits the sphere. OK before the spheres move because this tension is pointing straight upwards. Now recall again, the sum of the forces is equal to the mass multiplied by the acceleration. This is Newton's second law. The forces we have acting are the tension force tee acting upwards. We have the force of gravity which is given by M G acting downwards. So we have T minus M G is equal to the mass M multiplied by AC. OK. Now I'm writing this acceleration as a centripetal acceleration since this sphere is suspend suspended. OK. Then it's going to be rotating at that length of the string. We have this in some triple acceleration pointing towards that string or in the direction of the string. All right, now, we have an acceleration in our equation. But what we're really interested is interested in is the speed OK? That speed that will provoke a string crack. So let's recall that we can write a centripetal acceleration ac as V squared over R. So our equation becomes T minus M G is equal to M V squared divided by R. All right. Now let's substitute in the values. We know we know that this tension forces five newtons. We were given that in the problem that's that breaking strength of the spring or the string. Sorry. So we wanna include that force because we have to, if we use the breaking force of the string, that's gonna give us the speed that that occurs at OK. The mass, we have an 18 g dart, we wanna convert this to kilograms. So to go from grams to kilograms, we multiply by 10 to the exponent negative three mm. We're gonna add that to the mass of the ball as well. OK. Again, this is the sphere immediately as that dart tip hits it. OK. That's our free body diagram. And so the mass is the total mass of the sphere dart combination. So we have the mass of the dart plus the mass of the sphere g. And again, multiplying by 10 to the exponent negative three to convert to kilograms, we multiply that by the acceleration due to gravity 9. m per second squared, this is equal to, again, that mass 18 times 10 to the exponent negative three kg plus and the same thing 360 times 10 to the exponent negative three kg. That's gonna give us 0.36 kg. I'm gonna write it like that on this side. So we save some space multiplied by the velocity V squared divided by the radius. OK? And the radius of this centripetal motion is gonna be 0.75 m the length of that strength. OK. So now the only unknown we have in this equation is V let's go ahead and solve for V. So on the left hand side, we get five newtons minus 3. 818 newtons. And I've written 9.8 m per second squared. Let's use 9.81. We'll use a little bit more precision here. So we have 9.81 m per second squared for the acceleration due to gravity. How we get newtons? We have kilograms multiplied by meters per second squared. OK? And that is equivalent to a Newton. On the right hand side we're gonna simplify those masses divide by 0.75 m and we get 0.504 kg per meter multiplied by V squared. We divide by that 0.504. We get that V squared is equal to 2.563145. We had newtons kilogram meter per second squared. We divided by kilogram per meter which leaves us with meters squared per second squared. And we're gonna take the square root. Let me get that V is equal to 1.601 m per second. And when we take the square route, we get the positive and negative route. What we're worried about here is speed. So we don't care about the direction. So we're just gonna take the positive route here to look at speed. OK? So we found this speed of 1.601 m per second is that the speed we were looking for? It's actually not OK. This is the speed of that sphere dark combination immediately after the collision that will cause a string crack. OK? In other words, this is V F from our momentum equation. OK. This is the final velocity right after that collision. So what we found was that our V F max is equal to 1.601 m per second. And we know that V F is equal to one divided by 21 V not D. OK. This tells us that in that initial speed of the dart must be 21 multiplied by that final speed. We found that that final speed is 1. m per second. Ok? That maximum final speed and this is equal to 33.6 m per second. Ok. So that is that velocity we were looking for the initial velocity of the dart right before the collision. Ok. Let's walk through this one more time. We found what we were looking for. But this was a long problem. OK. We wanted to find the maximum speed of the dart just before the collision so that this string would not crack. The first thing we did was use conservation of momentum to figure out the relationship between the final speed right after the collision of the dart and sphere combination in terms of that initial velocity of the dart before the collision. OK. So we wrote down that relationship, we use the forces equation to find out the maximum final speed after the collision that we can have without invoking a string crack. And then we substituted that back into the equation. We found to see what speed that relates to for our initial dart velocity. All right. So the correct answer here is 33.6 m per second, which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

A 20 kg wood ball hangs from a 2.0-m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A 1.0 kg projectile traveling horizontally hits and embeds itself in the wood ball. What is the greatest speed this projectile can have without causing the wire to break?

Key Concepts

Tension in a Wire

Conservation of Momentum

Centripetal Force

Watch next