Question

suppose you push a hockey puck of mass m across frictionless ice for a time 1.0 s, starting from rest, giving the puck speed v after traveling distance d. you repeat the experiment with a puck of mass 2m.

Answer 1

Answer:

jajahankqjqhabanaijanaakajhqnakoajdjkauhnwkqijanmaaoiajajakakiajksjdijajjsijwnjsisjnsnsksjjwnskskjskskiwjsjwjaiiajdjnenihdnnwoiakksdijdkoalkakajjdijwakkaisjsjkskoaiakskosksjakskks

Answer 2

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.

I hope this is understandable and if my answer helped you please can you give me a brainliest its will really help me and thank you have a nice day!!