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m Δ v = m [ 0 ( v 1 j ^ ) ] m Δ v = + m v 1 j ^ .

We can get the speed of the phone just before it hits the floor using either kinematics or conservation of energy. We’ll use conservation of energy here; you should re-do this part of the problem using kinematics and prove that you get the same answer.

First, define the zero of potential energy to be located at the floor. Conservation of energy then gives us:

E i = E 1 K i + U i = K 1 + U 1 1 2 m v i 2 + m g h drop = 1 2 m v 1 2 + m g h floor .

Defining h floor = 0 and using v i = ( 0 m/s ) j ^ gives

1 2 m v 1 2 = m g h drop v 1 = ± 2 g h drop .

Because v 1 is a vector magnitude, it must be positive. Thus, m Δ v = m v 1 = m 2 g h drop . Inserting this result into the expression for force gives

F = Δ p Δ t = m Δ v Δ t = + m v 1 j ^ Δ t = m 2 g h Δ t j ^ .

Finally, we need to estimate the collision time. One common way to estimate a collision time is to calculate how long the object would take to travel its own length. The phone is moving at 5.4 m/s just before it hits the floor, and it is 0.14 m long, giving an estimated collision time of 0.026 s. Inserting the given numbers, we obtain

F = ( 0.172 kg ) 2 ( 9.8 m/s 2 ) ( 1.5 m ) 0.026 s j ^ = ( 36 N ) j ^ .

Significance

The iPhone itself weighs just ( 0.172 kg ) ( 9.81 m/s 2 ) = 1.68 N ; the force the floor applies to it is therefore over 20 times its weight.

Check Your Understanding What if we had assumed the phone did bounce on impact? Would this have increased the force on the iPhone, decreased it, or made no difference?

If the phone bounces up with approximately the same initial speed as its impact speed, the change in momentum of the phone will be Δ p = m Δ v ( m Δ v ) = 2 m Δ v . This is twice the momentum change than when the phone does not bounce, so the impulse-momentum theorem tells us that more force must be applied to the phone.

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Momentum and force

In [link] , we obtained an important relationship:

F ave = Δ p Δ t .

In words, the average force applied to an object is equal to the change of the momentum that the force causes, divided by the time interval over which this change of momentum occurs. This relationship is very useful in situations where the collision time Δ t is small, but measureable; typical values would be 1/10th of a second, or even one thousandth of a second. Car crashes, punting a football, or collisions of subatomic particles would meet this criterion.

For a continuously changing momentum—due to a continuously changing force—this becomes a powerful conceptual tool. In the limit Δ t d t , [link] becomes

F = d p d t .

This says that the rate of change of the system’s momentum (implying that momentum is a function of time) is exactly equal to the net applied force (also, in general, a function of time). This is, in fact, Newton’s second law, written in terms of momentum rather than acceleration. This is the relationship Newton himself presented in his Principia Mathematica (although he called it “quantity of motion” rather than “momentum”).

If the mass of the system remains constant, [link] reduces to the more familiar form of Newton’s second law. We can see this by substituting the definition of momentum:

F = d ( m v ) d t = m d v d t = m a .

The assumption of constant mass allowed us to pull m out of the derivative. If the mass is not constant, we cannot use this form of the second law, but instead must start from [link] . Thus, one advantage to expressing force in terms of changing momentum is that it allows for the mass of the system to change, as well as the velocity; this is a concept we’ll explore when we study the motion of rockets.

Practice Key Terms 2

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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