Question

A2100 KGtruck travelling north at 41 km/h turns east and accelerates to 51 km/h. (a) What is the change in the truck’s kinetic energy? What are the (b) Magnitude and (c) Direction of the change in its momentum?

Short answer

(a) The change in kinetic energy of the truck is $K=7.5\times {10}^{4}J$.

(b) The magnitude of the change in momentum of the truck is $\left|\overrightarrow{P}\right|=3.8\times {10}^{4}kg.m/s$.

(c) The direction of the change in momentum of the truck is $\theta =39°$.

Step-by-step solution

Listing the given quantities:    

The mass of the truck is m = 2100 kg .

The initial velocity of the truck is$v_{1y}=41km/h=11.38m/s$.

The final velocity of the truck is $v_{2y}=51km/h=14.16m/s$.

Understanding the concept of law of conservation of momentum:

The law of conservation of momentum states this. For two or more bodies in an isolated system acting on each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed.

You can use the concept of the law of conservation of momentum and change of kinetic energy to find the answers.

Formulae:

Write the equation for kinetic energy as below.

$K=\frac{1}{2}m\left(v_f^2-v_i^2\right)$

The momentum is defined by,

$\left|\overrightarrow{p}\right|=m\left|\overrightarrow{v}\right|$

The direction is calculated by using the following formula.

$\tan \theta =\frac{v_y}{v_x}$

Here, $\theta$ is the angle.

(a) Calculations for change in kinetic energy of truck:

The change in kinetic energy of the truck is,

$$\begin{gathered} K=\frac{1}{2}m\left(v_f^2-v_i^2\right) \\ K=\frac{1}{2}m\left(v_{2x}^2-v_{1y}^2\right) \\ =\frac{1}{2}\times 2100kg-\times \left[{\left(14.16m/s\right)}^2-\left(11.38m/s^2\right)\right] \\ =7.5\times {10}^{4}J \end{gathered}$$

Hence, the change in kinetic energy of the truck is K =$7.5\times {10}^{4}J$.

(b) Calculations for change in magnitude of momentum of truck:

The velocity vector of the truck is,

$$\begin{gathered} \overrightarrow{v}=\overrightarrow{v_x}+\overrightarrow{v_y} \\ =\overrightarrow{v_{2x}}+\left(-\overrightarrow{v_{1y}}\right) \end{gathered}$$

The magnitude of the change in velocity is,

$\left|\overrightarrow{v}\right|=\sqrt{{\left(\overrightarrow{v_{2x}}\right)}^2+{\left(-\overrightarrow{v_{1y}}\right)}^2}$

Substitute known values in the above equation.

$$\begin{gathered} \left|\overrightarrow{v}\right|=\sqrt{{\left(14.16m/s\right)}^2+{\left(-11.38m/s\right)}^2} \\ =\sqrt{\left(200.5056+129.5044\right)m^2/s^2} \\ =\sqrt{\left(330.01\right)m^2/s^2} \\ =18.16m/s \end{gathered}$$

The magnitude of the change in momentum is,

$$\begin{gathered} \left|\overrightarrow{P}\right|=m\left|\overrightarrow{v}\right| \\ =2100kg\times 18.16m/s \\ =3.8\times {10}^{4}kg.m/s \end{gathered}$$

Hence, the magnitude of the change in momentum of the truck is $3.8\times {10}^{4}kg.m/s$.

(c) Calculations for direction of change in momentum of truck:

The direction of the change in momentum:

$$\begin{gathered} \tan \theta =\frac{v_y}{v_x} \\ \theta ={\tan }^{-1}\left(\frac{-v_{1y}}{v_{2x}}\right) \\ ={\tan }^{-1}\left(\frac{11.38m/s}{14.16m/s}\right) \\ =39° \end{gathered}$$

Hence, the direction of the change in momentum of the truck is $39°$.