Question Video: Finding the Change in the Momentum of a Body Descending in Water Mathematics

A sphere of mass 83 g fall vertically from a height of 8.1 m onto a section of horizontal ground. It rebounded and reached a height of 3.6 m. Given that the duration of impact was 0.42 seconds and the acceleration due to gravity is 9.8 m/s², find the average impact force to the nearest 2 decimal places.

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Video Transcript

A sphere of mass 83 grams fall vertically from a height of 8.1 meters onto a section of horizontal ground. It rebounded and reached a height of 3.6 meters. Given that the duration of impact was 0.42 seconds and the acceleration due to gravity is 9.8 meters per second squared, find the average impact force to the nearest two decimal places.

So in this question, we’re thinking about a sphere that falls to the ground from a height of 8.1 meters. We’re told that the sphere rebounds, bouncing back up to a new height of 3.6 meters. We’ve also been told the mass of the sphere, the duration of the impact, and the acceleration due to gravity, and we’re being asked to calculate the average impact force. Now, when we think about an impact force, we generally think about the force exerted by some moving object on the thing that it collides with. So in this question, we could think of the impact force as being the force that the sphere exerts on the ground during the short amount of time that they’re in contact.

However, Newton’s third law tells us that every action has an equal and opposite reaction, which means that whenever the sphere is exerting a force on the ground, the ground is exerting an equal and opposite force on the sphere. And in this question, it’s this force that we’re going to think of as the impact force. We’re told in the question that the duration of impact is 0.42 seconds, so we know that there’s an impact force for 0.42 seconds and that this force is sufficient to change the motion of the sphere from downward motion to upward motion. Specifically, we want to find the average value of this force over the 0.42 seconds.

Now this question is pretty complicated as the motion of the sphere is constantly changing. First, the sphere is falling, so it undergoes constant acceleration due to gravity. Then, it collides with the ground, during which time some unknown force is acting on it for 0.42 seconds. And finally, the sphere rebounds, which means again it’s accelerating downwards due to gravity. Only this time it’s been given some upward velocity. Now, despite the relatively complicated motion of the sphere, our answer to this question depends on one pretty simple equation, 𝐼 equals 𝐹 times Δ𝑡. In other words, the impulse 𝐼 experienced by an object is equal to the force 𝐹 acting on that object multiplied by the time Δ𝑡 for which that force acts.

Now, since we’re interested in finding the force that acts on the sphere, we can rearrange this expression to make 𝐹 the subject by dividing both sides by Δ𝑡, giving us 𝐹 equals 𝐼 over Δ𝑡. Recall that an impulse is equivalent to a change in momentum. That means that we can rewrite this expression as 𝐹 equals Δ𝑝 over Δ𝑡, where Δ𝑝 is the change in momentum of the sphere. This equation forms the basis of our answer. If we can find the change in momentum of the sphere as it bounces and divide this by the duration of the impact, then this will tell us the average force acting on the sphere during the impact. Since we’ve been given Δ𝑡, the impact time, we just need to figure out Δ𝑝.

Now, just to be clear, since we’re interested in the average force acting on the sphere during the impact, that means the change in momentum Δ𝑝 that we’re interested in is the change in the momentum of the sphere that occurs during the impact. That means we’re interested in the change in momentum that occurs between the moment the object hits the ground, at which point it’s traveling downward, and the moment the object bounces off the ground, at which point it’s traveling upward. So if we say that when the sphere hits the ground, its momentum is given by 𝑝 subscript hit and when it bounces off the ground, its momentum is given by 𝑝 subscript bounce, then we can say that the change in momentum Δ𝑝 is equal to 𝑝 bounce minus 𝑝 hit.

Now, since an object’s momentum 𝑝 is given by its mass 𝑚 multiplied by its velocity 𝑣 and we’re given the mass of the sphere in the question, then in order to calculate the momentum of the sphere as it hits the ground and as it bounces off, we need to calculate its velocity at these moments. Fortunately, it is possible to calculate these velocities using a single kinematic equation, 𝑣 squared equals 𝑢 squared plus two 𝑎𝑠. This tells us that for an object undergoing constant acceleration 𝑎, its final velocity squared is equal to its initial velocity squared plus two times its acceleration times its displacement. We’re going to apply this equation twice: once to work out the velocity of the sphere as it hits the ground, we’ll call this 𝑣 hit, and once to calculate its velocity just as it bounces off the ground, which we’ll call 𝑣 bounce.

Now, at this point, we should note that most of the quantities we’re dealing with in this question are vector quantities. This includes momentum, force, final velocity, initial velocity, acceleration, and displacement. Since all of these quantities are vectors, this means they can take positive or negative values depending on which direction they’re acting. So we need to take care to define which direction is our positive direction and which direction is our negative direction. So, for this question, let’s say that any vector pointing upward is positive and any vector pointing downward is negative.

Okay, so now let’s use this equation to calculate the velocity of the object as it hits the ground. To do this, we need to think about the first part of the sphere’s motion, which is when it falls to the ground from a height of 8.1 meters. Thinking about this part of the sphere’s motion, let’s consider what values these variables take. 𝑣 is the object’s final velocity. So in this case, that’s the velocity of the sphere as it hits the ground. This is the quantity we’re trying to work out, and we’ve given it the name 𝑣 hit. Next is 𝑢. This is the initial velocity of the object. In this case, this is zero meters per second since we assume that the sphere falls from rest.

Next is the acceleration of the object 𝑎. Since the object is falling under gravity and acceleration due to gravity is given in the question as 9.8 meters per second squared, 𝑎 takes a value of negative 9.8 meters per second squared because this acceleration acts in the negative direction. Finally, 𝑠 is the displacement of the sphere during this fall. Since it falls from a height of 8.1 meters to the ground, it’s displaced 8.1 meters in the negative direction. So 𝑠 takes a value of negative 8.1 meters. Substituting all of these values into the equation gives us the expression 𝑣 hit squared equals zero squared plus two times negative 9.8 times negative 8.1, which works out as 158.76. Since this is the value of 𝑣 hit squared, we know that 𝑣 hit is equal to the square root of 158.76, which is 12.6.

One thing that we should be careful of here is that when we take the square root of a number, it actually has two answers, as a positive square root and a negative square root. So we could say that the square root of 158.76 is positive 12.6 or negative 12.6. Now, when we’re dealing with physical problems, for example, the sphere bouncing off the ground, we often find that we need to choose either the positive or the negative solution, depending on what we know about the situation. In this case, we know that 𝑣 hit is acting downwards since the sphere is falling. Because we’ve already decided that any vector acting downward is negative, this means that we know 𝑣 hit is negative. So in this case, we choose the negative square root. So 𝑣 hit is equal to negative 12.6 meters per second.

So now that we found the velocity of the object just as it hits the ground, we can calculate its momentum at this point, 𝑝 hit, by multiplying 𝑣 hit by the mass of the sphere 𝑚. And we’re told in the question that the mass of the sphere is 83 grams, which in standard units is 0.083 kilograms. This means that 𝑝 hit is equal to 0.083 times negative 12.6, giving us a momentum of negative 1.0458 kilogram meters per second. So let’s make a note of this over here. And now let’s use the same kinematic equation to calculate the velocity of the sphere as it bounces off the ground. This time, we’re interested in the values that these variables take during the part of the sphere’s motion where it travels upward from the ground to a height of 3.6 meters.

In this case, 𝑣, the final velocity of the sphere, is equal to zero meters per second. This is because the velocity of the sphere is instantaneously zero when it reaches its maximum height. The initial velocity of the sphere 𝑢 during this motion is the velocity at which it leaves the ground and we’ve defined this as 𝑣 bounce. The acceleration of the sphere is the same as before. Even though this time the sphere is traveling upward, it’s still accelerating downward due to gravity. So, once again, 𝑎 takes a value of negative 9.8 meters per second squared. And finally, since the object travels upward from the ground to a height of 3.6 meters, the displacement 𝑠 is positive 3.6 meters. Substituting these values into the equation again gives us zero squared equals 𝑣 bounce squared plus two times negative 9.8 times 3.6.

Now we’re trying to solve this for 𝑣 bounce. So we can start by calculating the value of two times negative 9.8 times 3.6, which is negative 70.56. And we can then add 70.56 to each side of the equation to give us 𝑣 bounce squared equals 70.56. Next, we can square root both sides of the equation, which tells us that 𝑣 bounce is equal to the square root of 70.56, which is 8.4. Once again, because we obtained this velocity by taking the square root of a number, we can choose either the positive or the negative value depending on what we know about the question. In this case, since we know that 𝑣 bounce is a velocity in the positive direction, we choose the solution positive 8.4 and the units for this are meters per second.

Now we have the velocity of the sphere at this point, we can calculate the corresponding momentum by multiplying the velocity of 8.4 meters per second by the mass of the sphere 0.083 kilograms, giving us a value for the momentum of the sphere just after it bounces of 0.6972 kilogram meters per second. Okay, so now we found the momentum of the sphere just before it hits the ground and just after it bounces off the ground. We can now calculate the change in momentum Δ𝑝 that took place during the impact. 𝑝 bounce minus 𝑝 hit is equal to 0.6972 minus negative 1.0458, which gives us an overall change in momentum of 1.743 kilogram meters per second. And here we can see why it was important that we kept track of which vectors were positive and which vectors were negative. We could easily have got a different answer if we got the sign of either of these values wrong.

Anyway, now that we found the change in the sphere’s momentum during the impact, we can substitute this value into this equation along with the duration of the impact Δ𝑡 in order to tell us the average impact force. 1.743 kilogram meters per second divided by 0.42 seconds gives us a final answer of 4.15 newtons. And double-checking the question, we can see that we’ve already given our answer to the correct number of decimal places. So if a sphere of mass 83 grams falls vertically from a height of 8.1 meters then rebounds and reaches a height of 3.6 meters, then given the duration of the impact is 0.42 seconds, we can calculate the average impact force is 4.15 newtons.

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