91Ó°ÊÓ

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, $\stackrel{→}{F}is\left(12.0-3.00t^2\right)\hat{\mathrm{i}}$.

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.

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

(a)

The impulse – linear momentum theorem gives

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

Hence the impulse can be calculated as

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

(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

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