### 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.