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 a

Question

11 days ago

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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 a

fter 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?

Physics

2 answers:

Keith_Richards [1K] · Answer 1 · 2026-02-22T13:42:13+03:00

Scenario 1:

Let the applied force in both scenarios be "F"

where m denotes the mass of the puck

the puck's acceleration is defined as

a = F/m

t = duration of the push = 1 second

d = distance covered

v = velocity attained

v₀ = starting speed = 0 m/s

Utilizing the formula,

v = v₀ + at

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

v = F/m

The distance covered is calculated as

d = v₀t + (0.5)at²

d = (0)(1) + (0.5)(F/m)(1)²

d = F/(2m)

Scenario 2:

a)

When the puck’s mass becomes "2m"

the acceleration then is

a' = F/(2m)

t'= duration of push =?

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

v' = achieved speed = v = F/m

v'₀ = initial speed = v₀ = 0 m/s

Using the formula,

v' = v'₀ + a' t'

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

t' = 2 seconds

b)

The distance covered is represented as

d' = v'₀t' + (0.5)a't'²

F/(2m) = (0)t' + (0.5)(F/(2m))t'²

t' = 2 seconds

Softa [913] · Answer 2 · 2026-02-22T13:44:05+03:00

According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)