**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 \(5.19\;s\).