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Momentum & Impulse : Impulse
 

[ momentum ][ impulse ][ units ][ vector problems ]

 

 

 

 

Momentum

Momentum is by definition the product of mass and velocity. So strictly speaking momentum is a vector quantity.

momentum = mass(kg) x velocity(ms-1)

Hence the unit of momentum is (kg.ms-1).

Impulse of a force

This is simply the force multiplied by the time the force acts.

We can obtain an expression for this in terms of momentum from Newton's Second Law equation F=ma, where the force F is constant.

Remembering that velocity, force and therefore impulse are vector quantities.

For a mass m being accelerated by a constant force F, where the impulse is J , v1 is initial velocity and v2 is final velocity:

Ft = m(v2- v1)

J = Ft

J = m(v2- v1)

 

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Units

Since impulse is the product of force and time, the units of impulse are (Newtons) x (seconds), or N s .

 

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Vector problems

Vector type questions on impulse are solved by first calculating the change in momentum. This gives a vector expression for the impulse. Using Pythagoras, the magnitude of the impulse can then be found. The anglular direction is calculated from the coefficients of unit vectors i and j.

Example #1

A particle of mass 0.5 kg moves with a constant velocity of    (3i + 5j) m.s-1 .
After being given an impulse, the particle then moves off with a constant velocity of (2i - 3j) m.s-1 .

Calculate:

i)   the impulse
ii)  the magnitude of the impulse( to 2 d.p.)
iii) the direction of the impulse(θ degrees to the x-axis)

i)

v1= (3i + 5j)      v2 = (2i - 3j)     m = 0.5 kg

using J = m(v2- v1)

J = 0.5(2i - 3j) - 0.5(3i + 5j)

J = i - 1.5j - 1.5i - 2.5j

J = (1 - 1.5)i + (-1.5 - 2.5)j

J = (0.5i - 4j) N.s

ii)

magnitude of impulse = √ [( 0.5)2 + (-4)2 ] = √ [16.23]

    = 4.03 N.s

iii) direction tan-1 θ = (4)/(0.5) = 8

θ = 82.8749o = 82.87o (2 d.p.) clockwise to the x-axis

 

Example #2

A particle of mass 2.5 kg is moving with a constant velocity of    (2i + j) m.s-1 .
After an impulse, the particle moves with a constant velocity of (4i + 3j) m.s-1 .

Calculate:

i)   the impulse
ii)  the magnitude of the impulse( to 2 d.p.)
iii) the angle(θo) the impulse makes with the x-axis

v1= (2i + j)     v2 = (4i + 3j)     m = 2.5 kg

using J = m(v2- v1)

J = 2.5(4i +3j) - 2.5(2i + j)

J = 10i + 7.5j - 5i - 2.5j

J = (10 - 5)i + (7.5 - 2.5)j

J = (5i + 5j)

ii)

magnitude of impulse = √ [( 5)2 + (5)2 ] = √ [50] = 7.07 N.s

iii) direction tan-1 θ = (5)/(5) = 1

θ = 45o anticlockwise to the x-axis

 

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