Chapter 9, Problem 90A

a.

To determine

The change in momentum of the dragster if a $845\text{kg}$ dragster accelerates from rest to $100.0\text{km/h}$ in $0.90\text{s}$ .

Answer to Problem 90A

  $2.347\times {10}^4\text{kg}\cdot \text{m/s}$

Explanation of Solution

Given:

The mass of the dragster, $m=845\text{kg}$

Initial velocity of the dragster, $v_i=0\text{m/s}$

Final velocity of the dragster, $v_f=100\text{km/h }\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}}=27.78\text{m/s}$

Time taken, $\Delta t=0.90\text{s}$

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

  $\begin{array}{c}{p}_f-p_i=m{v}_f-m{v}_i\\ \Delta p=m\left(v_f-v_i\right)\\ =845\left(27.78-0\right)\\ =2.347\times {10}^4\text{kg}\cdot \text{m/s}\end{array}$

Conclusion:

Thus, the change in momentum of the dragster is $2.347\times {10}^4\text{kg}\cdot \text{m/s}$ .

b.

To determine

The average force exerted on the dragster.

Answer to Problem 90A

  $2.6078\times {10}^4\text{N}$

Explanation of Solution

The mass of the dragster, $m=845\text{kg}$

Initial velocity of the dragster, $v_i=0\text{m/s}$

Final velocity of the dragster, $v_f=100\text{km/h }\times \frac{1000\text{m}}{1\text{km}}\times \frac{1\text{h}}{3600\text{s}}=27.78\text{m/s}$

Time taken, $\Delta t=0.90\text{s}$

From part (a), the change in momentum of the dragster, $\Delta p=2.347\times {10}^4\text{kg}\cdot \text{m/s}$

Formula used:

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

  $p=mv$

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:

Then, using the given and calculated values in above formula, the magnitude of the force applied on the dragster will be,

  $\begin{array}{c}F\Delta t=\Delta p\\ F\left(0.9\right)=2.347\times {10}^4\text{kg}\cdot \text{m/s}\\ F=2.6078\times {10}^4\text{N}\end{array}$

Conclusion:

Thus, the magnitude of the force on the dragster is $2.6078\times {10}^4\text{N}$ .

c.

To determine

What exerts the above force in dragster.

Answer to Problem 90A

Track through the friction.

Explanation of Solution

Introduction:

A $845\text{kg}$ dragster on a race track accelerates from rest to $100.0\text{km/h}$ in $0.90\text{s}$ .

When a dragster moves on a race track, it experiences friction on the track due to which a force is exerted on it resulting in its change of momentum.

Conclusion:

Thus, the track through friction exerts a force on the dragster.