91Ó°ÊÓ

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 $v_{2y}=51km/h=14.16m/s$.

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,

$\left|\stackrel{→}{p}\right|=m\left|\stackrel{→}{v}\right|$

The direction is calculated by using the following formula.

$\tan \mathrm{θ}=\frac{v_y}{v_x}$

Here, $\mathrm{θ}$ is the angle.

The change in kinetic energy of the truck is,

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

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

The velocity vector of the truck is,

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

The magnitude of the change in velocity is,

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

Substitute known values in the above equation.

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

The magnitude of the change in momentum is,

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

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

The direction of the change in momentum:

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

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