Question

A 0.160-kg hockey puck is moving on an icy, frictionless, horizontal surface. At t=0 the puck is moving to the right at 3.02 m/s .

part A)

Calculate the magnitude of the velocity of the puck after a force of 25.4 N directed to the right has been applied for 6.0×10⁻² s

V= m/s

Answer 1

Let us consider the right side direction as positive and the left side direction as negative.

Mass of the hockey puck = m = 0.16 kg

Initial velocity of the hockey puck = V₁ = 3.02 m/s

Final velocity of the hockey puck = V₂

Force on the puck = F = 25.4 N

Time period the force is applied for = T = 6 x 10^-2 sec = 0.06 sec

Impulse of the force on the hockey puck = I

I = FT

I = (25.4)(0.06)

I = 1.524 N.s

Impulse is equal to the change in momentum of the object.

I = mV₂ - mV₁

1.524 = (0.16)V₂ - (0.16)(3.02)

V₂ = 12.5 m/s

A) Magnitude of velocity of the hockey puck after the force has been applied = 12.5 m/s