Set 4 Problem number 1

An object of mass 15 kg experiences a net force of 330 Newtons for .6 seconds.

Use Newton's Second Law and your knowledge of uniformly accelerated motion to find its change in velocity.

Use the Impulse-Momentum Theorem to obtain the same result.

Solution

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

The acceleration of the object will be 330 Newtons / 15 kg = 22 m/s^2.

During the .6 seconds the velocity of the object will therefore change by

`dv = 22 m/s^2 * .6 seconds = 13.2 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 = 330 Newtons * .6 seconds = 198 Newton seconds = 198 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 = 198 kg m/s / ( 15 kg) = 13.2 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: