Question

A fan cart of mass 0.8 kginitially has a velocity of (0.9,0,0)m/s. Then the fan is turned on, and the air exerts a constant force of (-0.3,0,0)Non the cart for 1.5s. (a) What is the change in momentum of the fan cart over this 1.5 sinterval? (b) What is the change in kinetic energy of the fan cart over this 1.5 sinterval?

Short answer

(a)The change in the momentum of the fan cart is (0.225,0,0)kg.m/s .

(b) The change in the kinetic energy of the fan cart is (-0.2788,0,0)J .

Step-by-step solution

Identification of the given data

The given data is listed below as,

• The mass of the fan cart is,m=0.8 kg
• The initial velocity of the fan cart is,u=(0.9,0,0)m/s
• The force exerted by air on the fan cart is,F=(-0.3,0,0)N
• The time for which the force is exerted by air is,t=1.5s

Significance of the equation of motion

The equation of motion mainly describes the physical system’s behavior in terms of the functions of the time.

The first equation of motion is expressed as follows,

$v-u=at$

Here,$v$ is the final velocity of the object,$u$ is the initial velocity of the object,$a$ is the acceleration and$t$ is the time taken.

(a) Determination of the change in the momentum

The equation of the force can be expressed as:

$F=m.a$

Here,$m$is the mass of the cart, and$a$is the acceleration of the cart.

Rearrange the above equation.

$a=\frac{F}{m}$

The expression for the first equation of the motion for the fan cart is expressed as:

$$\begin{gathered} v=u+at \\ =u+\frac{F}{m}t \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} v=\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\frac{\left(-0.3,0,0\right)\mathrm{N}}{0.8\mathrm{kg}}1.5\mathrm{s} \\ =\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\left(-0.375,0,0\right)\mathrm{N}/\mathrm{kg}.1.5\mathrm{s} \\ =\left(0.9,0,0\right)\mathrm{m}/\mathrm{s}+\left(-0.5625,0,0\right)\mathrm{N}.\mathrm{s}/\mathrm{kg}\times \frac{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}} \\ =\left(-0.3375,0,0\right)\mathrm{m}/\mathrm{s} \end{gathered}$$

The equation of the change in the momentum is expressed as:

$∆p=p_2-p_1$

Here,$p_2$is the final momentum and $p_1$is the initial momentum

The above equation can be expanded as:

$$\begin{gathered} ∆p=\left(mv\right)-\left(mu\right) \\ =m\left(v-u\right) \end{gathered}$$

Substitute all the values in the above expression.

role="math" localid="1657035004902" $$\begin{gathered} ∆p=0.8kg\times \left(\left(\left(0.3375,0,0\right)m/s\right)-\left(\left(0.9,0,0\right)m/s\right)\right) \\ =0.4kg\times \left(\left(-0.5625,0,0\right)m/s\right) \\ =\left(-0.225,0,0\right)kg.m/s \end{gathered}$$

Thus, the change in the momentum of the fan cart is (-0.225,0,0)kg.m/s .

(b) Determination of the change in the kinetic energy

The equation of the change in the kinetic energy is expressed as:

$∆K.E.=K{E}_2-K{E}_1$

Here, $K{E}_2$is the final kinetic energy and $K{E}_1$is the initial kinetic energy.

The above equation can be expanded as:

$$\begin{gathered} ∆KE=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2 \\ =\frac{1}{2}m\left(v^2-u^2\right) \end{gathered}$$

Substitute all the values in the above expression.

$$\begin{gathered} ∆K.E.=\frac{1}{2}\times 0.8kg\times \left({\left(\left(0.3375,0,0\right)m/s\right)}^2-{\left(\left(0.9,0,0\right)m/s\right)}^2\right) \\ =0.4kg\times \left(\left(0.113,0,0\right)m^2/s^2-\left(0,81,0,0\right)m^2/s^2\right) \\ =0.4kg\times \left(\left(-0.697,0,0\right)m^2/s^2\right) \\ =\left(\left(-0.2788,0,0\right)kg.m^2/s^2\times \frac{1J}{1kg.m^2{s}^2}\right) \\ =\left(0.2788,0,0\right)J \end{gathered}$$

Thus, the change in the kinetic energy of the fan cart is (-0.2788,0,0)J .