9.2 Impulse and collisions  (Page 5/10)

The iphone drop

Strategy

We need to make a couple of reasonable estimates, as well as find technical data on the phone itself. First, let’s suppose that the phone is most often dropped from about chest height on an average-height person. Second, assume that it is dropped from rest, that is, with an initial vertical velocity of zero. Finally, we assume that the phone bounces very little—the height of its bounce is assumed to be negligible.

Solution

The impulse $\overrightarrow{J}$ equals the change in momentum,

so

Next, the change of momentum is

We need to be careful with the velocities here; this is the change of velocity due to the collision with the floor. But the phone also has an initial drop velocity [ ${\overrightarrow{v}}_{\text{i}}=\left(0\text{m/s}\right)\widehat{j}$ ], so we label our velocities. Let:

• ${\overrightarrow{v}}_{\text{i}}=$ the initial velocity with which the phone was dropped (zero, in this example)
• ${\overrightarrow{v}}_1=$ the velocity the phone had the instant just before it hit the floor
• ${\overrightarrow{v}}_2=$ the final velocity of the phone as a result of hitting the floor

[link] shows the velocities at each of these points in the phone’s trajectory.

With these definitions, the change of momentum of the phone during the collision with the floor is

Since we assume the phone doesn’t bounce at all when it hits the floor (or at least, the bounce height is negligible), then ${\overrightarrow{v}}_2$ is zero, so