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.

Explanation:

Let a Force pushes the hockey puck of mass m.

Then acceleration, $a= \frac{F}{m}$

From the equation of motion,

$\Rightarrow v=u+at\\ v=0+\frac{F}{m}\Delta t$ ......(1)

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

$a'=\frac{F}{2m}$

and t' is the time taken.

The final speed is v,

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

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

Answer 2

V = AT

v = a/2 T

d = 1/2 a T^2
d = 1/2 a/2 T^2

1/2 at^2  = 1/4 aT^2
2t^2 = T^2

T = (Sqrt2)t

Hope this helps