# An object with a mass of 1.5 kg changes its velocity +15 m/s during a time interval of 3.5 seconds what impulse was delivered to the objects

QUESTION POSTED AT 01/06/2020 - 02:41 PM

QUESTION POSTED AT 01/06/2020 - 02:41 PM

The answer is 10.5 kg m/s

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So:

a = Δv/Δt ⇒ F = m · a = m · Δv/Δt

F = m · Δv/Δt ⇒ I = m · Δv/Δt · Δt

Since Δt can be cancelled out, impulse can be expressed as:

I = m · Δv = m · (v2 - v1)

It is given:

m = 1.5 kg

v1 = 15 m/s

v2 = 22 m/s

I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.

Impulse (I) is the multiplication of force (F) and time interval (Δt): I = F · Δt

Force (F) is the multiplication of mass (m) and acceleration (a): F = m · a

Acceleration (a) can be expressed as change in velocity (v) divided by time interval (Δt): a = Δv/Δt

So:

a = Δv/Δt ⇒ F = m · a = m · Δv/Δt

F = m · Δv/Δt ⇒ I = m · Δv/Δt · Δt

Since Δt can be cancelled out, impulse can be expressed as:

I = m · Δv = m · (v2 - v1)

It is given:

m = 1.5 kg

v1 = 15 m/s

v2 = 22 m/s

I = 1.5 · (22 - 15) = 1.5 · 7 = 10.5 kgm/s.

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