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.
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?

Answer 1

Case 1 :

let the force of push in both cases be "F"

m = mass of puck

acceleration of the puck is given as

a = F/m

t = time of push = 1 sec

d = distance traveled

v = speed gained

v₀ = initial speed = 0 m/s

Using the equation ,

v = v₀ + a t

v = 0 + (F/m) (1)

v = F/m

distance traveled is given as

d =  v₀ t + (0.5) a t²

d = (0)(1) + (0.5) (F/m) (1)²

d = F/(2m)

Case 2 :

a)

when the mass of puck is "2m"

acceleration is given as

a' = F/(2m)

t'= time of push = ?

d' = distance traveled = d = F/(2m)

v' = speed gained = v =  F/m

v'₀ = initial speed = v₀ = 0 m/s

Using the equation ,

v' = v'₀ + a' t'

(F/m) = 0 + (F/(2m)) t'

t' = 2 sec

b)

distance traveled is given as

d' =  v'₀ t' + (0.5) a' t'²

F/(2m) = (0) t'  + (0.5) (F/(2m)) t' ²

t' = 2 sec

Answer 2

Newton's second law ...Force = momentum change/time.momentum change = Forcextme.also, F=ma -> a=F/m - the more familiar form of Newton's second law
using one of the kinematic equations for m ...  V=u+at; u=0; a=F/m -> V=(F/m)xt.-> t=mV/F using one of the kinematic equations for 2m ... V=u+at; u=0; a=F/2m -> V=(F/2m)xt. -> t=2mV/F (twice as long, maybe ?)
I think I've made a mistake somewhere below, but I think that the principle is right ...using one of the kinematic equations for m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/m; t=1;  -> d=(1/2)(F/m)=F/2musing one of the kinematic equations for 2m ...  s=ut + (1/2)at^2); s=d;u=0;a=F/2m; t=1;  -> d=(1/2)(F/2m)=F/4m (half as far ????? WHAT ???)