2015 AP Physics 1 free response 4

Two identical spheres are released from a device at time equals zero from the same height ( h ), as shown above. Sphere A has no initial velocity and falls straight down. Sphere B is given an initial horizontal velocity of magnitude ( v_0 ) and travels a horizontal distance ( d ) before it reaches the ground. The spheres reach the ground at the same time ( t_f ), even though sphere B has more distance to cover before landing. Air resistance is negligible.

The dots below represent spheres A and B. Draw a free body diagram showing and labeling the forces, not components, exerted on each sphere at time ( t_f / 2 ).

So, we can see our spheres here. When this little device releases, sphere A goes straight down. Sphere B will go, well, it's vertical in the vertical direction; it'll go down just the same way. It'll be accelerated in just the same way as sphere A, but it has some horizontal velocity, and that makes it move out and hit the ground ( d ) to the right. When it hits the ground, that's ( t_f ).

When they're up here, this is at ( t = 0 ) when they're released, and then this is at ( t = t_f ). They say a free body diagram at ( t_f / 2 ). While both of them are in flight, the only force acting on each of them is just going to be the force of gravity. Since the spheres are identical, the force of gravity is going to be identical; they have the same mass.

Let me draw. So that right over there is the force of gravity on sphere A, and that is the force of gravity on sphere B. We could write "force of gravity" or "force of gravity." If we want, we could say the magnitude is ( f_g ); if we want ( f_g ) or we could label it as ( m ) times the gravitational field. This is equal to ( m ) times the gravitational field, and that's it. While they're mid-flight, the only force acting on them, we're assuming air resistance is negligible, is the force of gravity, which is going to be the same because they have the same mass; they're identical spheres.

All right, let's tackle the next part of this. On the axis below, sketch and label a graph of the horizontal components of the velocity of sphere A and sphere B as a function of time.

All right, I'll do sphere A first. This is pretty straightforward. Sphere A, if you will remember, sphere A has no horizontal velocity the entire time. We're talking about it only; it's only going to be accelerated in the vertical direction. It's going to be accelerated downwards. So sphere A has no horizontal velocity the entire time. Now, sphere B is going to be a little bit more interesting, slightly more interesting. Its velocity, they tell us that its initial velocity is ( v_0 ), its initial horizontal velocity, I should say, has a magnitude of ( v_0 ).

Since air resistance is negligible, it's going to continue going to the right at ( v_0 ) until it hits the ground. So sphere B, if this is, and I'm just going to pick one of these as ( v_0 ); let's say that this right over here is ( v_0 ), that's the magnitude of its horizontal velocity. Well, sphere B is going to be at that velocity. Actually, let me just make it a little clearer; it's going to be at that velocity until it hits the ground, until ( t_f ). So this is from ( t = 0 ) to ( t_f ).

The entire time while the sphere is in the air, its horizontal component of its velocity is just going to be constant. It's not going to be slowed down by anything because we're assuming air resistance is negligible. Then right when it hits the ground, if you think about the force that is stopping it, it's essentially friction, but then it very quickly goes down to a velocity of a magnitude of horizontal velocity of zero.

All right, now let's tackle the last part of this. Now you could label this if you want; let me actually label it. This is sphere B, and this is sphere A right over there. In sphere B, if you want, you could show it overwriting sphere A, so your ( v_B ) would be zero after that; it's not continuing to move on to the right, or at least they don't tell us anything about that.

Finally, in a clear, coherent paragraph-length response, explain why the spheres reach the ground at the same time, even though they travel different distances. Include references in your answers to parts A and B.

All right, so let me think about it. I'll try to write a clear, coherent paragraph-length response. From ( t = 0 ) to ( t = t_f ), the only force acting on the spheres is the downward force of gravity. At ( t = 0 ), they both have zero vertical velocity; the magnitude of the velocity in the vertical direction is zero for both of them. After ( t = 0 ), they are accelerated at the same rate. Their vertical components of velocity are always the same, and they have the same vertical distance to cover. So they hit the ground at the same time.

Let me make sure that makes sense. After ( t = 0 ), they are accelerated at the same rate, so their vertical components of velocity are always the same. Since they have the same vertical distance to cover, they will hit the ground at the same time. They do have different horizontal velocities, but that does not affect the time in which they cover the same vertical distance. You could write something to that effect, and you could also write that if you were to add the components of sphere B's velocities, it would actually have a larger velocity if you were not thinking in either the horizontal or the vertical direction.

So it does indeed cover more distance in space over the same amount of time, but if you think about it just in the vertical direction, it's covering the same distance in the same time. At any given point in time in the vertical direction, it actually has the same velocity; it's being accelerated in the same way and starts off at a magnitude of velocity of zero.