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