Question

A fancart of mass 0.8 kg initially has a velocity of < 0.8, 0, 0 > m/s. Then the fan is turned on, and the air exerts a constant force of < −0.3, 0, 0 > N on the cart for 1.5 seconds. What is the change in momentum of the fancart over this 1.5 second interval?

Answer 1

To solve this problem we will apply the concepts related to the change of the momentum. At the same time we will use Newton's second law to obtain the acceleration of the object and the kinematic equations of linear motion.

Newton's second law is defined as,

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

Here,

F = Force

m = Mass

a = Acceleration

Our values are given as,

$v_i = 0.8 m/s$

$v_f = ?$

$t = 1.5s$

$F = -0.3N$

According to this values we have that the acceleration is,

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

$a = \frac{-0.3}{0.8}$

$a = -0.375m/s^2$

Now applying the concepts of kinematic equations we have that the final velocity is,

$v_f = v_i + at$

$v_f = 0.8m/s + (-0.375m/s^2)(1.5s)$

$v_f = 0.3375m/s$

By definition we know that momentum is the product between mass and velocity, so the change in momentum would be given by

$\Delta p = p_f - p_i$

$\Delta p = mv_f-mv_i$

$\Delta p = m(v_f-v_i)$

$\Delta p = (0.8)(0.3375-0.9)$

$\Delta p = - 0.45kg \cdot m/s$

Therefore the change in momentum of the fancart is -0.45kg m/s