a 5500-kg freight trunk accelerates from 4.2 m/s to 7.8 m/s in 15.0 s by the application of a constant force a) what change in momentum occurs
b) how large of a force is exerted?

Answer :

(a) The change in momentum of the trunk is 19800 kgm/s,

(b) The force exerted on the trunk is 3960 N

Change in momentum:

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

(a) To calculate the change in momentum of the trunk, we use the formula below.

Formula:

  • ΔM = m(v-u)............... Equation 1

Whee:

  • ΔM = change in momentum of the trunk
  • m = mass of the trunk
  • v = final velocity of the trunk
  • u = initial velocity of the trunk.

From the question,

Given:

  • m = 5500 kg
  • v = 7.8 m/s
  • u = 4.2 m/s

Substitute these values into equation 1

  • ΔM = 5500(7.8-4.2)
  • ΔM = 5500(3.6)
  • ΔM = 19800 kgm/s.

(b) To calculate how large is the force was exerted on the trunk, we use the formula below.

Formula:

  • F = ΔM/t................. Equation 2

Where:

  • F = Force exerted on the trunk
  • t = time.

From the question,

Given:

  • t = 15.0 s.

Substitute into equation 2

  • F = 19800/5
  • F = 3960 N.

Hence, (a) The change in momentum of the trunk is 19800 kgm/s, (b) The force exerted on the trunk is 3960 N

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