Impulse

When an object collides with another object, transfer of energy takes place. This energy comes from the force exerted by the object to the other object to which it collides. A force applied to an object causes it to accelerate as in the Newton's 2nd Law of Motion, 

\(F=ma\)                                 (1) 

Expressing acceleration as the change in velocity per change of time, (1) will become

\(F=m\; \frac{\Delta v}{\Delta t}\)                                (2)

Take note that \(m\Delta v\) is the change in momentum of an object, thus, (2) can be expressed as

\(F=\frac{\Delta p}{\Delta t}\)                                    (3)

and the change in momentum is 

\(F\Delta t=\Delta p\)                                 (4)

where \(F\Delta t\) is a quantity called impulse. Equation (4) describes impulse as the force needed to change the momentum of an object in a given unit of time. The equation also shows that the impulse is equal to the change in momentum of an object.


Example 1.

Solve for the force needed by a baseball player to catch a \(141.7\; g\) baseball in \(5\;\text{seconds}\) if it is moving at \(49.2 \;m/s\)

Given:

\(m=141.7\;g\\ v=49.2\;m/s\\ t=5\;s\)


Solution 1.

Convert the mass to kilograms.

\(141.7\;g\times\frac {1\;kg}{1000\;g}=0.1417\;kg\).


Solution 2.

Solve for the force of the player by using the equation for impulse \(F\Delta t=m\Delta v\)

\(F=\frac {m\Delta v}{\Delta t}=\frac {(0.1417\;kg)(49.2\;m/s)}{5\;s}=1.39\;N\)

Therefore, the player needed \(1.39\;N\) of force to catch the ball.


Example 2.

A \(135-N\) force is applied to a \(35-kg\) object. How much time is needed to increase the speed of the object from \(25\;m/s\) to \(45\;m/s\)?

Given:

\(F=135\;N\\ m=35\;kg\\ v_i=25\;m/s\\ v_f=45\;m/s\)


Solution:

Use \(\Delta v=v_f-v_i\) in the equation. Using \(F\Delta t=m\Delta v\), the formula for time is

\(t=\frac{m(v_f-v_i)}{F}=\frac{35\;kg(45\;m/s-25\;m/s)}{135\;N}=\frac{700\;kg\cdot m/s}{135\;N}=5.19\;s\)

thus, the time needed to change the speed of the object is \(5.19\;s\).