Suppose you push a hockey puck of mass m across frictionless ice for 1.0 s, starting from rest, giving the puck speed v after traveling distance d. If you repeat the experiment with a puck of mass 2m, pushing with the same force, a. How long will you have to push for the puck to reach the same speed v?

b. How long will you have to push for the puck to travel the same distance d?

The kinematic equation for motion along a straight line is as follows:$x$-axis ,

$v_{1x}=v_{0x}+a_{x}t$

We $v_{1x}$is the highest possible velocity$v_{0x}$is the beginning speed,$a_x$is the acceleration and $t$is the time interval.

The following is the kinematics expression for distance travelled:

$x_1-x_0=+v_{01}t+\frac{1}{2}{a}_x{t}^2$

By all above $x_1$is the final position and $x_0$initial position of the puck.

For mass puck $A$, the formulation of Newton's second rule of motion$\left(m\right)$is as follows,

$F_x=m{a}_x$

Rearrange the phrase of Newton's second law of motion to find the value of acceleration.$\left(a_x\right)$,

$a_x=\frac{F_x}{m}$

The time it takes for the puck to reach the specified velocity$v$is calculated as follows,

Calculation of the time required by mass puck A$m$to reach the speed $v$is as follows,

Putting$v$for $v_{1x},0\mathrm{m}/\mathrm{s}$for $v_{0x},\frac{F_x}{m}$for $a_x$and $t_{\mathrm{m}p\mathrm{x}}$for $t$in the kinematics equation,

$v=0\mathrm{m}/\mathrm{s}+\left(\frac{F_x}{m}\right)t_{\mathrm{m},\text{puk}}$

$t_{\mathrm{m},\text{puck}}$&$=\frac{mv}{F_x}$

For mass puck B, the formulation of Newton's second equation of motion$(2m)$is as follows,

$F_x=2m{a}_x$

To calculate the value of acceleration, rearrange the phrase of Newton's second law of motion$\left(a_x\right)$,

$a_x=\frac{F_x}{2m}$

The amount of time it takes for the puck to attain the desired speed $v$is calculated as follows,

Calculation of the time required by mass puck B$2m$to reach the speed $v$is as follows,

Putting$v$for$v_{1x},0\mathrm{m}/\mathrm{s}$for $v_{0x},\frac{F_x}{2m}$for $a_x$and $t_{\text{2m puck}}$for $t$in the kinematics equation,

$v=0\mathrm{m}/\mathrm{s}+\left(\frac{F_x}{2m}\right)t_{2\mathrm{m}\text{pudk}}$

$t_{2\mathrm{m}\text{pokk}}=\frac{2mv}{F_x}$

$=2\left(\frac{mv}{F_x}\right)$

Putting $t_{\text{m.puck}}$for $\frac{mv}{F_x}$from the expression of time of puck A of mass $m$,

$t_{2\mathrm{m},\text{pack}}=2t_{\mathrm{m},\text{pok}}$

As a result, puck B takes twice as long to reach the same speed as puck mass$A$.

Calculate the amount of time puck$A$ of mass takes to travel$m$is as follows,

Putting$d$for $x_1-x_0,0\mathrm{m}/\mathrm{s}$for $v_{0x},\frac{F_x}{m}$for $a_x$and$t_{\mathrm{m}\text{pat}}$for $t$ in the kinematics equation,

$d=0+\frac{1}{2}\left(\frac{F_x}{m}\right){\left(t_{\mathrm{m},\text{puok}}\right)}^2$

$t_{\mathrm{mpuak}}=\sqrt{\frac{2md}{F_x}}$

$\text{Calculation of time taken by puck}B\text{of}$

Calculation of the time required by mass puck B$m$is as follows, Putting$v$for $v_{1x},0\mathrm{m}/\mathrm{s}$for $v_{0x},\frac{F_x}{2m}$for $a_x$and $t_{2\mathrm{m}\text{, pus}}$for $t$in the kinematics equation,

$d=0+\frac{1}{2}\left(\frac{F_x}{2m}\right){\left(t_{2\mathrm{m},\text{puck}}\right)}^2$

$t_{2\mathrm{m},\text{puck}}=\sqrt{\frac{4md}{F_x}}$

$=\sqrt{2}\left(\sqrt{\frac{2md}{F_x}}\right)$

Putting $t_{\mathrm{m},\text{puck}}$for $\sqrt{\frac{2md}{F_x}}$from the mass puck A statement of time $m$,

$t_{2\mathrm{m},\text{puck}}=\sqrt{2}\left(t_{\mathrm{m},\text{puck}}\right)$

As a result, the puck spends more time on the ice$\mathbf{B}$ to travel a considerable distance $d$is $\sqrt{2}$times and time taken by the puck's mass $A$.