### Video Transcript

A bullet with a mass of 10 grams is
travelling at 450 metres per second and hits a stationary toy car of mass 400
grams. The bullet comes to rest 0.01
seconds after hitting the car. What force is required to bring the
bullet to rest in 0.01 seconds? What is the acceleration of the car
due to the impact from the bullet?

Okay, so in this question, what
we’ve been told is that initially we have a bullet of mass 10 grams moving towards a
car at 450 metres per second. And that car is a toy car which has
a mass of 400 grams and it’s initially stationary. Now, at some point, the bullet and
the car collide and that results in the bullet coming to rest in 0.01 seconds. Now, in the first part of the
question, we’ve been asked to try and work out the force required to do this, to
completely stop the bullet in 0.01 seconds if it was initially travelling at 450
metres per second.

Now, to answer this question, we
need to recall Newton’s second law of motion. This law tells us that the net
force on an object 𝐹 subscript net is equal to the rate of change of the object’s
momentum or in other words the change in momentum of the object divided by the time
taken for this momentum change to occur. Now, in this case, we know that the
bullet was initially travelling at 450 metres per second. And then, it stops when it collides
with the car. So we know that the bullet’s
velocity is changing; it’s decreasing. And because that velocity is
changing, we know that the momentum of the bullet is changing because remember the
momentum of an object is equal to the mass of the object multiplied by its
velocity. And so, the change in momentum of
the object — which we’ll call Δ𝑝 — is going to be equal to the change in mass
multiplied by velocity of the object.

Now, in the situation where the
mass of the object remains constant as it does for the bullet, we know that it stays
at 10 grams throughout the collision and after the collision as well, in that case
we can say that the change in momentum of the object is equal to that constant mass
multiplied by the change in velocity of the object. So let’s go about finding the
change in momentum of the bullet. We know that it’s equal to the mass
of the bullet which is 10 grams or in base units, it’s 0.01 kilograms. And then, we need to multiply this
mass by the change in velocity of the bullet. And the change in velocity of the
bullet is the final velocity minus the initial velocity.

Now we know that the final velocity
is zero metres per second because the bullet is stationary at the end. And the initial velocity is 450
metres per second because that’s what it was travelling at before. And now, note that we’ve made the
assumption that the bullet’s initial velocity to the right was positive and hence
our change in velocity is going to be negative because the bullet is decelerating;
it’s losing speed. But anyway, so we find that the
change in momentum of the bullet is equal to negative 4.50 kilograms metres per
second.

And so now that we’ve calculated
this change in momentum, we can work out the net force on the bullet which is the
force that’s required to bring the bullet to rest in 0.01 seconds. To do this, we simply need to say
that the net force on the bullet 𝐹 subscript net is equal to the change in momentum
which we’ve calculated as negative 4.50 kilograms metres per second divided by the
time over which the collision occurs which we’ve been told is 0.01 seconds. And then, because we’re working in
base units — so that’s kilograms metres per second for momentum and seconds for time
— we know that the final unit of this calculation is going to be kilograms metres
per second squared, which is equivalent to the newton, the unit of force.

So when we evaluate the right-hand
side of this equation, we find that the net force on the bullet is negative 450
newtons. Now because we’ve simply been asked
to find the force required and haven’t specifically been asked to worry about
direction, we’ll omit the negative sign in our answer. We’ll just say that the magnitude
of the force required to decelerate the bullet in 0.01 seconds so that it comes to
rest is 450 newtons. So now that we’ve calculated that,
let’s look at the next part of the question.

What is the acceleration of the car
due to the impact from the bullet? Well, to answer this, we can recall
Newton’s third law of motion. Newton’s third law tells us that if
an object — let’s say object A — exerts a force on another object, object B. Then, B exerts an equal and
opposite force on A. Now, we’ve seen in this collision
that there was a force of 450 newtons exerted on the bullet in order to decelerate
it. Well, that force would have been
exerted by the car onto the bullet due to the collision that occurred between the
two objects. And so, by Newton’s third law of
motion, we can see that the bullet will exert a force in the opposite direction onto
the car. And the magnitude of this force is
the same, 450 newtons.

And then, we can recall that for an
object with a constant mass as we’ve seen already, the change in momentum of the
object is equal to that constant mass multiplied by the change in velocity. But then, if we take that equation
for the change in momentum and substitute it back into Newton’s second law of
motion, we see that the net force on an object 𝐹 subscript net is equal to the mass
multiplied by the change in velocity divided by the change in time or the time taken
for, in this case, the collision to occur. And then, we can notice that the
change in velocity divided by the change in time is equal to the acceleration of the
object. And so, in the situation where an
object’s mass remains constant, Newton’s second law reduces to the net force is
equal to the mass of the object multiplied by its acceleration.

Now, we already know the force
acting on the car; it’s 450 newtons. So we know the net force. And we also know the mass of the
car. We know that it’s 400 grams. Hence, we can work out the
acceleration by rearranging this equation. We can say that the acceleration of
the car is equal to the net force divided by its mass. And that ends up being 450 newtons
divided by 400 grams. Or in base units, the denominator
becomes 0.4 kilograms. And then evaluating the right-hand
side, we find that the acceleration of the car is 1125 metres per second
squared. And at this point, we found the
answer to our question.