Question

FIGURE EX11.6 is an incomplete momentum bar chart for a collision that lasts $10ms$. What are the magnitude and direction of the average collision force exerted on the object?

Short answer

The magnitude is $F=800N$and the direction is $-x$-axis

Step-by-step solution

Given Information 

We have given chart for a collision that lasts $10ms$.

Simplify

Any condition when an object receives a force for a short time, this force with time is called the impulse. So it is the area under the curve of the force-versus-time graph and it is the same as momentum. The impulse is the quantity $J_x$and we a force deliver an impulse to an object with an initial momentum $P_{ix}$it changes its momentum to $P_{ix}$and it is given in equation (11.9) in the form

$P_{fx}=P_{ix}+J_x$ (1)

A momentum bar chart is similar to an energy bar chart which shows the momentum principle. The initial momentum from the bar chart is localid="1649332374797" $P_{ix}=6kg\times m/s$and the final momentum is given in the negative direction by $P_{fx}=-2kg\times m/s$. We solve equation(1) for $J_x$and plug the values for localid="1649332479972" $P_{fx}$and $P_{ix}$to get $J_x$

$$\begin{gathered} J_x=P_{fx}-P_{ix} \\ =-2kg\times m/s-6kg\times m/s \\ =-8kg\times m/s \end{gathered}$$

The impulse is the quantity $J_x$and it is given in equation (11.6) in the form

$J_x=F∆t$

Where localid="1649332845745" $∆t=10ms=10\times {10}^{-3}s$. We solve this equation for $F$and putting the value for $J_x$and $∆t$, we will get $F$as

$$\begin{gathered} F=\frac{J_x}{∆t} \\ =\frac{-8kg\times m/s}{10\times {10}^{-3}s} \\ =-800N \end{gathered}$$

The negative sign means the force is delivered in the opposite direction of the force is$-x$-axis