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

(a) The time taken for the puck to travel the same speed is 2.0 s.

(b) The time taken for the puck to travel the same distance is 2.0 s.

Applied force

The force applied to push the hockey is calculated as follows;

F = ma

where;

• m is the mass
• a is the acceleration

When the mass doubles

The acceleration of the mass is calculated as follows;

$F = (2m) a\\\\a = \frac{F}{2m}$

Since the acceleration of motion reduces by 2, the time of motion will double = 2 (1.0 s) = 2.0 s.

Also, the time taken for the puck to travel the same distance is 2.0 s.

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