Set 4 Problem number 1

A net force of 1118 Newtons is applied to an object of mass 13 kg for .3 seconds.

Find its change in velocity using Newton's Second Law and your knowledge of uniformly accelerated motion, and also using the Impulse-Momentum Theorem.

Solution

First we find the change in velocity using the acceleration and time interval:

The acceleration of the object will be 1118 Newtons / 13 kg = 86 m/s^2.

During the .3 seconds the velocity of the object will therefore change by

`dv = 86 m/s^2 * .3 seconds = 25.8 m/s.

Next we find the change in velocity using the impulse and change in momentum:

The impulse of the force is the product F * `dt = 1118 Newtons * .3 seconds = 335.4 Newton seconds = 335.4 kg m / s.

This is equal to the change in momentum, by the Impulse-Momentum Theorem.

Since the mass is unchanging the change in momentum is m `dv.

The change in velocity is therefore obtained by dividing the change in momentum by the mass:

`dv = 335.4 kg m/s / ( 13 kg) = 25.8 m/s.

Generalized Solution

If a force F acts on object of constant mass m for `dt seconds, the object will experience acceleration a = F / m for `dt seconds, resulting in velocity change

`dv = a `dt = (F / m) `dt = (F `dt) / m.

When the relationship `dv = (F `dt) / m is rearranged into the form

m `dv = F `dt

we have the Impulse-Momentum Theorem for object of constant mass.

We can use the Impulse-Momentum Theorem to find any of the quantities `dv, F, m or `dt given the values of three of these quantities.

University Physics Notes:

Using calculus we can prove that d ( mv ) = F dt, which applies to situations in which mass and/or velocity vary.

Since d ( mv ) = m dv + v dm, we have

m dv + v dm = F dt.

When mass is constant, dm = 0 so v dm disappears and we have m dv = F dt, as before.
When velocity is constant we have v dm = F dt, which tells us that to maintain a changing mass at constant velocity requires a force which depends on how quickly the mass is changing (i.e., F = v dm/dt).

Explanation in terms of Figure(s), Extension

Figure description: