The driver accelerates a 330.0 kg snowmobile, which results in a force being exerted that speeds up the snowmobile from 6.00 m/s to 23.0 m/s over a time interval of 60.0 s. a. What is the snowmobile's change in momentum? What is the impulse on the snowmobile? b. What is the magnitude of the average force that is exerted on the snowmobile?

(a) 5610 kgm/s

(b) 5610 Ns.

(c) 78. 64 N

Explanation:

a. Change in momentum: This can be defined as the product of the mass of a body to its change in velocity. The S. I unit of change in momentum is kgm/s.

Mathematically, change in momentum is expressed as

ΔM = mΔv ... Equation 1

Where ΔM = change in momentum, m = mass of snowmobile, Δv = change in velocity.

Given: m = 330 kg, Δv = v₂-v₁ = 23-6 = 17 m/s.

Note: v₁ and v₂ are the initial and the final velocity of the snowmobile.

ΔM = 330 (17)

ΔM = 5610 kgm/s.

(b) Impulse: This can be defined as the product and force and time. The S. I unit of impulse is Ns.

Note: From Newton's second law of motion, impulse is equal to change in momentum.

Therefore,

I = ΔM ... Equation 2

Where I = impulse of the force.

Since ΔM = 5610 kgm/s.

Therefore

I = 5610 Ns.

Thus the impulse = 5610 Ns.

(c) Force: This can be defined as the product of the mass of a body and its acceleration. The S. I unit of force is Newton (N).

F = ma ... Equation 3

F = force, m = mass of the body, a = acceleration

But,

a = (v₂-v₁) / t

Where v₂ = 23.0 m/s, v₁ = 6.0 m/s t = 60.0 s.

a = (23-6) / 60

a = 0.283 m/s².

Substituting the value a and m into equation 3

F = 330 (0.2383)

F = 78.639 N.

F ≈ 78. 64 N