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11.8 Momentum

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Momentum

Momentum is a physical quantity which is closely related to forces. Momentum is a property which applies to moving objects.

Momentum

Momentum is the tendency of an object to continue to move in its direction of travel. Momentum is calculated from the product of the mass and velocity of an object.

The momentum (symbol p ) of an object of mass m moving at velocity v is:

p = m · v

According to this equation, momentum is related to both the mass and velocity of an object. A small car travelling at the same velocity as a big truck will have a smaller momentum than the truck. The smaller the mass, the smaller the velocity.

A car travelling at 120 km · hr - 1 will have a bigger momentum than the same car travelling at 60 km · hr - 1 . Momentum is also related to velocity; the smaller the velocity, the smaller the momentum.

Different objects can also have the same momentum, for example a car travelling slowly can have the same momentum as a motor cycle travelling relatively fast. We can easily demonstrate this. Consider a car of mass 1 000 kg with a velocity of 8 m · s - 1 (about 30 km · hr - 1 ). The momentum of the car is therefore

p = m · v = ( 1000 kg ) ( 8 m · s - 1 ) = 8000 kg · m · s - 1

Now consider a motor cycle of mass 250 kg travelling at 32 m · s - 1 (about 115 km · hr - 1 ). The momentum of the motor cycle is:

p = m · v = ( 250 kg ) ( 32 m · s - 1 ) = 8000 kg · m · s - 1

Even though the motor cycle is considerably lighter than the car, the fact that the motor cycle is travelling much faster than the car means that the momentum of both vehicles is the same.

From the calculations above, you are able to derive the unit for momentum as kg · m · s - 1 .

Momentum is also vector quantity, because it is the product of a scalar ( m ) with a vector ( v ).

This means that whenever we calculate the momentum of an object, we need to include the direction of the momentum.

Khan academy video on momentum - 1

A soccer ball of mass 420 g is kicked at 20 m · s - 1 towards the goal post. Calculate the momentum of the ball.

  1. The question explicitly gives

    • the mass of the ball, and
    • the velocity of the ball

    The mass of the ball must be converted to SI units.

    420 g = 0 , 42 kg

    We are asked to calculate the momentum of the ball. From the definition of momentum,

    p = m · v

    we see that we need the mass and velocity of the ball, which we are given.

  2. We calculate the magnitude of the momentum of the ball,

    p = m · v = ( 0 , 42 kg ) ( 20 m · s - 1 ) = 8 , 4 kg · m · s - 1

  3. We quote the answer with the direction of motion included, p = 8,4 kg · m · s - 1 in the direction of the goal post.

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A cricket ball of mass 160 g is bowled at 40 m · s - 1 towards a batsman. Calculate the momentum of the cricket ball.

  1. The question explicitly gives

    • the mass of the ball ( m = 160 g = 0,16 kg), and
    • the velocity of the ball ( v = 40 m · s - 1 )

    To calculate the momentum we will use

    p = m · v

    .

  2. p = m · v = ( 0 , 16 kg ) ( 40 m · s - 1 ) = 6 , 4 kg · m · s - 1 = 6 , 4 kg · m · s - 1 in the direction of the batsman
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The Moon is 384 400 km away from the Earth and orbits the Earth in 27,3 days. If the Moon has a mass of 7,35 x 10 22 kg, what is the magnitude of its momentum if we assume a circular orbit?

  1. The question explicitly gives

    • the mass of the Moon (m = 7,35 x 10 22 kg)
    • the distance to the Moon (384 400 km = 384 400 000 m = 3,844 x 10 8 m)
    • the time for one orbit of the Moon (27,3 days = 27,3 x 24 x 60 x 60 = 2,36 x 10 6 s)

    We are asked to calculate only the magnitude of the momentum of the Moon (i.e. we do not need to specify a direction). In order to do this we require the mass and the magnitude of the velocity of the Moon, since

    p = m · v

  2. The magnitude of the average velocity is the same as the speed. Therefore:

    s = d Δ t

    We are given the time the Moon takes for one orbit but not how far it travels in that time. However, we can work this out from the distance to the Moon and the fact that the Moon has a circular orbit. Using the equation for the circumference, C , of a circle in terms of its radius, we can determine the distance travelled by the Moon in one orbit:

    C = 2 π r = 2 π ( 3 , 844 × 10 8 m ) = 2 , 42 × 10 9 m

    Combining the distance travelled by the Moon in an orbit and the time taken by the Moon to complete one orbit, we can determine themagnitude of the Moon's velocity or speed,

    s = d Δ t = C T = 2 , 42 × 10 9 m 2 , 36 × 10 6 s = 1 , 02 × 10 3 m · s - 1 .

  3. The magnitude of the Moon's momentum is:

    p = m · v = ( 7 , 35 × 10 22 kg ) ( 1 , 02 × 10 3 m · s - 1 ) = 7 , 50 × 10 25 kg · m · s - 1
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Vector nature of momentum

As we have said, momentum is a vector quantity. Since momentum is a vector, the techniques of vector addition discussed in Chapter  [link] must be used to calculate the total momentum of a system.

Two billiard balls roll towards each other. They each have a mass of 0,3 kg. Ball 1 is moving at v 1 = 1 m · s - 1 to the right, while ball 2 is moving at v 2 = 0 , 8 m · s - 1 to the left. Calculate the total momentum of the system.

  1. The question explicitly gives

    • the mass of each ball,
    • the velocity of ball 1, v 1 , and
    • the velocity of ball 2, v 2 ,

    all in the correct units!

    We are asked to calculate the total momentum of the system . In this example our system consists of two balls. To find the total momentum we must determine the momentum of each ball and add them.

    p t o t a l = p 1 + p 2

    Since ball 1 is moving to the right, its momentum is in this direction, while the second ball's momentum is directed towards the left.

    Thus, we are required to find the sum of two vectors acting along the same straight line. The algebraic method of vector addition introducedin Chapter  [link] can thus be used.

  2. Let us choose right as the positive direction, then obviously left is negative.

  3. The total momentum of the system is then the sum of the two momenta taking the directions of the velocities into account. Ball 1 is travellingat 1 m · s - 1 to the right or +1 m · s - 1 . Ball 2 is travelling at 0,8 m · s - 1 to the left or -0,8 m · s - 1 . Thus,

    p t o t a l = m 1 v 1 + m 2 v 2 = ( 0 , 3 kg ) ( + 1 m · s - 1 ) + ( 0 , 3 kg ) ( - 0 , 8 m · s - 1 ) = ( + 0 , 3 kg · m · s - 1 ) + ( - 0 , 24 kg · m · s - 1 ) = + 0 , 06 kg · m · s - 1 = 0 , 06 kg · m · s - 1 to the right

    In the last step the direction was added in words. Since the result in the second last line is positive, the total momentum of the systemis in the positive direction (i.e. to the right).

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Exercise

    1. The fastest recorded delivery for a cricket ball is 161,3 km · hr - 1 , bowled by Shoaib Akhtar of Pakistan during a match against England in the 2003 Cricket World Cup, held in South Africa. Calculate the ball's momentum if it has a mass of 160 g.
    2. The fastest tennis service by a man is 246,2 km · hr - 1 by Andy Roddick of the United States of America during a match in London in 2004. Calculate the ball's momentum if it has a mass of 58 g.
    3. The fastest server in the women's game is Venus Williams of the United States of America, who recorded a serve of 205 km · hr - 1 during a match in Switzerland in 1998. Calculate the ball's momentum if it has a mass of 58 g.
    4. If you had a choice of facing Shoaib, Andy or Venus and didn't want to get hurt, who would you choose based on the momentum of each ball.
  2. Two golf balls roll towards each other. They each have a mass of 100 g. Ball 1 is moving at v 1 = 2,4 m · s - 1 to the right, while ball 2 is moving at v 2 = 3 m · s - 1 to the left. Calculate the total momentum of the system.
  3. Two motor cycles are involved in a head on collision. Motorcycle A has a mass of 200 kg and was travelling at 120 km · hr - 1 south. Motor cycle B has a mass of 250 kg and was travelling north at 100 km · hr - 1 . A and B are about to collide. Calculate the momentum of the system before the collision takes place.

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Source:  OpenStax, Siyavula textbooks: grade 11 physical science. OpenStax CNX. Jul 29, 2011 Download for free at http://cnx.org/content/col11241/1.2
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