Video Transcript
A sphere of mass 125 grams was
moving along a section of horizontal ground at 165 centimeters per second when it
hit a vertical wall. Given that the wall applied an
impulse of magnitude 0.214 newton seconds to the sphere, find the speed of the
sphere as it bounced off the wall.
So we have a sphere moving along a
section of horizontal ground. And we’re told that it bounces off
a vertical wall. And we’re being asked to calculate
the speed of the sphere as it bounced off the wall. So we want to find out how fast the
sphere is travelling after it’s collided with the wall. The question tells us the mass of
the sphere is 125 grams. And we’re also told that the
initial speed of the sphere as it travels towards the wall is 165 centimeters per
second. And in addition to this, we’re told
that the wall applies an impulse of magnitude 0.214 newton seconds to the
sphere.
At this point, we can recall that
an impulse can be equated to a change in momentum. This means that when an object
experiences an impulse of a certain size, its momentum will change by that exact
amount. So when the question tells us that
the wall applies an impulse of magnitude 0.214 newton seconds to the sphere, what
this means is that the sphere’s momentum changes by 0.214 newton seconds during the
collision with the wall. So since this question tells us
information about the change in momentum that occurs during the collision, let’s
think about the momentum of the sphere before the collision with the wall and after
the collision with the wall.
Let’s recall that the momentum of
an object 𝑝 is given by that object’s mass 𝑚 multiplied by its velocity 𝑣. This means that the momentum of the
sphere before the collision, we could call this its initial momentum 𝑝 𝑖, is equal
to its mass, we’ll call this 𝑚 for now, multiplied by its velocity, we’ll call this
𝑣 𝑖 for initial velocity. Both of these quantities are given
in the question. We’re told that the mass of the
sphere is 125 grams and the speed of the sphere as it approaches the wall is 165
centimeters per second.
Now one thing we need to be careful
of here is that momentum, velocity, and even impulse are vector quantities. This means that they can take
positive or negative values depending on which way that they’re acting. So we need to decide which way is
our positive direction and which way is our negative direction. For this question, let’s say that
any vectors acting to the right are positive and any vectors acting to the left are
negative.
Because the initial velocity of the
sphere is going to the right, this means that 𝑣 𝑖 is pointing in the positive
direction, so it takes a positive value. So 𝑣 𝑖 is equal to positive 165
centimeters per second, which expressed in standard units is 1.65 meters per
second. We also know that the mass 𝑚 is
equal to 125 grams, which expressed in the standard units for mass is 0.125
kilograms.
Using these values, we can see that
the initial momentum of the sphere is equal to 0.125 kilograms multiplied by 1.65
meters per second, which gives us a value of 0.20625. Since we used standard units in our
calculation and we’re calculating momentum, this value takes the standard units for
momentum. Often, we express momentum in
kilogram meters per second. But we could equivalently use
newton seconds, which are the units used for impulse in the question.
Now that we’ve calculated the
initial value of the sphere’s momentum, we can use the information we’ve been given
about the impulse it receives to calculate its momentum after the collision. Let’s call this final momentum 𝑝
𝑓. To see exactly how we can do this,
we can rewrite this equation: Impulse is equal to a change in momentum. Using symbols, we could write this
as 𝐼 equals 𝛥𝑝. Now 𝛥𝑝, the change in momentum,
is equal to the difference between the final and initial values of momentum, which
we could write as 𝑝 𝑓 minus 𝑝 𝑖.
So if we rearrange this equation by
adding 𝑝 𝑖 to both sides, we can see that the final momentum of an object 𝑝 𝑓 is
equal to its initial momentum 𝑝 𝑖 plus the impulse 𝐼 that it receives. We can use this equation to
calculate the final momentum of the sphere. We just need to substitute in the
value of initial momentum, which we’ve already calculated, and a value for the
impulse 𝐼.
Now, at this point, we need to be
really careful to remember that impulse is a vector quantity and that the question
only tells us the magnitude of this quantity. The value that we use for 𝐼 might
be positive or negative depending on the direction that the impulse vector acts. Now because the sphere in this
question bounces off the wall, we know that its momentum changes direction. This is only possible if it
experiences an impulse that opposes its initial momentum. This means that the impulse acts
toward the left, which we’ve defined as the negative direction. So in this case, the impulse 𝐼
takes a value of negative 0.214 newton seconds.
Substituting both of these values
into this equation gives us 0.20625 minus 0.214, which is equal to negative 0.00775
newton seconds. And the fact that this value’s
negative signifies that the sphere is indeed moving to the left. Now all that’s left is to calculate
the velocity based on this momentum because momentum is equal to mass times
velocity, the final momentum 𝑝 𝑓 must be equal to mass times the final velocity 𝑣
𝑓.
We can rearrange this equation to
make 𝑣 𝑓 the subject by dividing both sides by 𝑚. This gives us 𝑣 𝑓 equals 𝑝 𝑓
over 𝑚. Substituting in our calculated
value of final momentum and the mass expressed in kilograms gives us negative
0.00775 divided by 0.125, which is equal to negative 0.062. The units here are the standard
units of velocity meters per second. Now because the question asks for
the speed of the sphere, our answer is given by the magnitude of this velocity, so
we can ignore the negative sign.
And the very final step is to
convert this speed from meters per second into the units of speed that are used in
the question, in this case, centimeters per second. And 0.062 meters per second is 6.2
centimeters per second. So there is our final answer. If a sphere of mass 125 grams was
moving along a section of horizontal ground at 165 centimeters per second then
bounces off a wall which applies an impulse of 0.214 newton seconds, the speed of
the sphere as it bounces off the wall is 6.2 centimeters per second.
