Video Transcript
Two spheres of masses 200 grams
and 350 grams were moving toward each other along the same horizontal straight
line. The first was moving at 14
meters per second and the second at three meters per second. The two spheres collided. As a result, the first sphere
rebounded at seven meters per second in the opposite direction. Given that the positive
direction is the direction of motion of the first sphere before the impact,
determine the impulse 𝐼 the second sphere exerted on the first one and the
speed 𝑣 of the second sphere after impact.
Now before we go any further,
we might observe that with our speeds, we’re working in meters per second,
whilst with the masses, we’re working in grams. So we’re going to convert their
masses to SI base units of mass, the kilogram. To do so, we divide each by
1000. We get 0.2 kilograms and 0.35
kilograms, respectively.
So how do we calculate the
impulse 𝐼 that the second sphere exerted on the first? Well, first, we know that the
magnitudes of the impulses on the spheres are equal. We also have information about
the velocity of the first sphere before and after the collision. This means we can calculate the
change in velocity of the first sphere due to the collision. If we call that the change in
𝑣 sub 𝑎, where 𝑎 is the name of the first sphere, then we can say that this
is equal to its final velocity minus its initial velocity.
Now, since it was initially
moving in the positive direction, it will rebound and move in the negative
direction. This means if it’s for
traveling at a speed of seven meters per second, its velocity is negative
seven. So the change in velocity is
negative seven minus 14. And that’s negative 21 meters
per second. Then, we can calculate the
impulse on the first sphere by finding its change in momentum. Momentum is of course mass
times velocity. So we could calculate the
momentum before and after and subtract them. Or alternatively, we can
multiply the momentum by the change in velocity. For our first sphere, we can
then say that its mass is 0.2 and its change in velocity is negative 21. So the impulse on this sphere
is 0.2 times negative 21, and that’s negative 4.2. And the units we use are newton
seconds.
Next, we know that an impulse
of equal magnitude, but opposite in sign, acts on the second sphere. In other words, the impulse on
the second sphere must be 4.2 newton seconds. But of course, we can link that
to its change in momentum. This means it’s equal to its
mass times its change in velocity. Now, since it was initially
moving in the negative direction at a speed of three meters per second, its
change in velocity is 𝑣 minus negative three. So its change in momentum is
0.35 times 𝑣 minus negative three. This right-hand side is
equivalent to 0.35 times 𝑣 plus three. And we calculated the impulse
on the second sphere was 4.2. So we can say 4.2 equals 0.35
times 𝑣 plus three.
Next, we divide through by
0.35. Now, 4.2 divided by 0.35 is
simply 12. Then, we subtract three from
both sides. 12 minus three is nine. So we’ve calculated that 𝑣 is
equal to nine meters per second. And so we’ve answered the
question. The impulse that the second
sphere exerted on the first, which we defined earlier to be 𝐽 sub 𝑎, is in
fact 𝐼 equals negative 4.2 newton seconds. Similarly, the speed 𝑣 of the
second sphere is the magnitude of nine meters per second, which is simply nine
meters per second.
