A 0.25 kg puck is initially stationary on an ice surface with negligible friction. At time t = 0, a horizontal force begins to move the puck. The force is given by$\overrightarrow{\mathbf{F}}\mathbf{=}\left(12.0-3.00t^2\right)\hat{\mathbf{i}}$, with in newton’s and tin seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between t = 0.500s and t =- 1.25 s? (b) What is the change in momentum of the puck between t = 0and the instant at which F = 0?

i) The mass of the puck, m is 0.25 kg.

ii) The initial speed of the puck, v is 0 m/s.

iii) The force acting on the puck, $\overrightarrow{F}is\left(12.0-3.00t^2\right)\hat{\mathrm{i}}$.

The force acting on the puck imparts impulse which is time dependent. The momentum change can be calculated using the impulse-linear momentum theorem stated as follows.

$$\begin{gathered} \mathbf{J}\mathbf{=}\mathbf{∆}\mathbf{p} \\ \mathbf{}\mathbf{}\mathbf{=}{\mathbf{\int }}_{{\mathbf{t}}_{\mathbf{1}}}^{{\mathbf{t}}_{\mathbf{2}}}\mathbf{Fdt} \end{gathered}$$

(a)

The impulse – linear momentum theorem gives

$J={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}Fdt$

Hence the impulse can be calculated as

$$\begin{gathered} J={\int }_{{\mathrm{t}}_1}^{{\mathrm{t}}_2}Fdt \\ ={\int }_{0.500}^{1.25}\left(12.0-3.00t^2\right)dt \\ ={\left[12.0t-3.00\frac{t^2}{3}\right]}_{0.500}^{1.25} \\ =\left[\left(12.0\times 1.25\right)-{\left(1.25\right)}^3\right]-\left[12.0\times 0.500-{\left(0.500\right)}^3\right] \\ =\left[\left(15.0-1.9531\right)-\left(6.00-0.125\right)\right] \\ =13.0469-5.875 \\ =7.17\mathrm{N}.\mathrm{s} \end{gathered}$$

(b)

First, we will find the time at which $F=0\mathrm{N}$

$$\begin{gathered} F=12.0-3.00t^2 \\ =0 \\ 3.00t^2=12.0 \\ {t}^2=4.00 \\ t=2.00\mathrm{or}-2.00 \end{gathered}$$

But time cannot be negative hence we discard t=-2.00 s

Hence t = + 2.00 s

Now we determine the change in momentum as

$$\begin{gathered} ∆P=J \\ ={\int }_{t_1}^{t_2}Fdt \\ ={\int }_{0.00}^{2.00}\left(12.0-3.00t^2\right)dt \\ ={\left[12.0t-3.00\frac{t^3}{3}\right]}_0^{2.00} \\ =\left[\left(12.0\times 2.00\right)-{\left(2.00\right)}^3\right]-0 \\ =24.0-8.00 \\ =16.0\mathrm{kg}.\mathrm{m}/\mathrm{s} \end{gathered}$$