Chapter 9, Problem 46A

a.

To determine

The change in momentum of a $5500\text{kg}$ truck if it accelerates from $4.2\text{m/s}$ to $7.8\text{m/s}$ in $15.0\text{s}$ by the application of a constant force.

Answer to Problem 46A

  $1.98\times {10}^4\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the truck, $m=5500\text{kg}$

Initial velocity of the truck, $v_i=4.2\text{m/s}$

Final velocity of the truck, $v_f=7.8\text{}\text{m/s}$

The time taken, $\Delta t=15.0\text{}\text{s}$

Formula used:

Momentum of an object of mass m and moving with velocity v is given by,

  $p=mv$

Calculation:

Using the given values in above formula, the change in momentum will be,

  $\begin{array}{c}\Delta p=p_f-p_f\\ =m\left(v_f-v_i\right)\\ =5500\left(7.8-4.2\right)\\ =1.98\times {10}^4\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the change in momentum of the truck is $1.98\times {10}^4\text{kg}\cdot \text{m/s}$ .

b.

To determine

The force exerted on a $5500\text{kg}$ truck if it accelerates from $4.2\text{m/s}$ to $7.8\text{m/s}$ in $15.0\text{s}$ by the application of a constant force.

Answer to Problem 46A

  $1320\text{N}$

Explanation of Solution

Given:

The mass of the truck, $m=5500\text{kg}$

Initial velocity of the truck, $v_i=4.2\text{m/s}$

Final velocity of the truck, $v_f=7.8\text{}\text{m/s}$

The time taken, $\Delta t=15.0\text{}\text{s}$

From part (a), change in momentum of the truck is $1.98\times {10}^4\text{kg}\cdot \text{m/s}$ .

Formula used:

Impulse on an object is equal to the change in momentum of an object,

  $\text{Impulse = }F\Delta t=p_f-p_i$

where $F$ is the applied force for time $\Delta t$ , $p_f$ is the final momentum, $p_i$ is the initial momentum.

Calculation:

Using the given and calculated values in above formula, the applied force will be,

  $\begin{array}{c}F\Delta t=\Delta p\\ F\left(15.0\right)=1.98\times {10}^4\\ F=1320\text{N}\end{array}$

Conclusion:

Thus, the force exerted on the truck is $1320\text{N}$ .