Question

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

Answer 1

Answer: Twice the previous time would be taken to reach the same speed v with the puck of mass 2m.

Step-by-step explanation:

Let a Force pushes the hockey puck of mass m.

Then acceleration,
`a= (F)/(m)`

From the equation of motion,

`\Rightarrow v=u+at\\ v=0+(F)/(m)\Delta t` ......(1)

In the second case, when mass is 2m, then acceleration,

`a'=(F)/(2m)`

and t' is the time taken.

The final speed is v,

`\Rightarrow v=0+ a't'\\ \Rightarrow \frac {F}{m}\Delta t=(F)/(2m)t'\\ \Rightarrow t'= 2\Delta t` using equation (1)

Hence, it would take two times the previous amount of time to push the pluck of double mass.