Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.
Explanation:
Let a Force pushes the hockey puck of mass m.
Then acceleration, a= \frac{F}{m}a=mF
From the equation of motion,
\begin{gathered}\➪ v=u+at\\ v=0+\frac{F}{m}\Delta t\end{gathered}⇒v=u+atv=0+mFΔt ......(1)
In the second case, when mass is 2m, then acceleration,
a'=\frac{F}{2m}a′=2mF
and t' is the time taken.
The final speed is v,
\begin{gathered}\➪ v=0+ a't'\\ \➪ \frac {F}{m}\Delta t=\frac{F}{2m}t'\\ \➪ t'= 2\Delta t\end{gathered}⇒v=0+a′t′⇒mFΔt=2mFt′⇒t′=2Δt using equation (1)
Hence, it would take two times the previous amount of time to push the pluck of double mass.
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