Question

A 0.18 kg puck is initially stationary on an ice surface with negligible friction. At time t = 0, a horizontal force begins to move the puck. The force is given by F→=(15.0-3.94t2)î, with F→ in newtons and t in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between t = 0.560 s and t = 1.79 s? (b) What is the change in momentum of the puck between t = 0 and the instant at which F = 0?

Answer 1

Mass of puck = m = 0.18 kg

Initial speed of puck = V₀ = 0 m/s

Force acting on puck = F

F = 15 - 3.94t²

Let t₀ be the time F becomes zero.

F = 0 = 15 - 3.94t₀²

t₀ = 1.95 sec

ma = 15 - 3.94t²

(0.18)a = 15 - 3.94t²

a = 83.33 - 21.88t²

Integrating we get,

V = 83.33t - 7.293t³ + C

At t=0 , V=0

Therefore C = 0.

V = 83.33t - 7.293t³

Speed of puck at t=0.56s = V₁

Speed of puck at t=1.79s = V₂

Speed of puck at t=1.95 that is when F=0 = V₃

V₁ = 83.33(0.56) - 7.293(0.56)³

V₁ = 45.38 m/s

V₂ = 83.33(1.79) - 7.293(1.79)³

V₂ = 107.33 m/s

V₃ = 83.33(1.95) - 7.293(1.95)³

V₃ = 108.41 m/s

Impulse on the puck from the force between t=0.56s and t=1.79s = I

Impulse is the change in momentum in the given time period.

I = mV₂ - mV₁

I = (0.18)(107.33) - (0.18)(45.38)

I = 11.151 N.s

Change in momentum between t=0 and the instant when F=0 = P

P = mV₃ - mV₀

P = (0.18)(108.41) - (0.18)(0)

P = 19.51 kg.m/s

a) Magnitude of impulse on puck between t=0.56s and t=1.79s = 11.151 N.s

b) Change in momentum of puck between t=0 and the instant at which F=0 = 19.51 kg.m/s