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9-33 a hockey puck of mass 150.0 g is sliding due east on a frictionless table with a speed of 10.0 m/s. suddenly, a constant force of magnitude 5.0 n and direction due north is applied to the puck for 1.5 s. find the momentum at the end of the 1.5s in component form.

Sagot :

The momentum at the end of the 1.5s of a hockey puck of mass 150.0 g is sliding due east with a speed of 10.0 m/s for northward and eastward component is 7.5 kg m / s and 1.5 kg m / s respectively.

For northward component,

Δp = F Δt

Δp = Change in momentum

F = Net force

Δt = Change in time

Δp = pf - pi

f = 5 N

Δt = 1.5 s

pi = 0 ( Since there is no momentum North )

pf = 5 * 1.5

pf = 7.5 kg m / s

For eastward component,

p = m v

m = Mass

v = Velocity

Since the force is acting perpendicular to velocity, there will be no change in velocity. So momentum remains constant for eastward component.

m = 150 g = 0.15 kg

v = 10 m / s

p = 0.15 * 10

p = 1.5 kg m / s

Therefore, the momentum of northward and eastward component is 7.5 kg m / s and 1.5 kg m / s respectively.

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