Page 109 - Physics Coursebook 2015 (A level)

A special case of Newton’s second law of motion
Imagine an object of constant mass m acted upon by a resultant force F. The force will change the momentum of the object. According to Newton’s second law of motion, we have:
F = Δp = mv − mu Δt t
where u is the initial velocity of the object, v is the final velocity of the object and t is the time taken for the change in velocity. The mass m of the object is a constant; hence the above equation can be rewritten as:
F = m(v − u) = m v − u tt
The term in brackets on the right-hand side is the acceleration a of the object. Therefore a special case of Newton’s second law is:
F = ma
We have already met this equation in Chapter 3. In Worked example 5, you could have determined the average force acting on the car using this simplified equation for Newton’s second law of motion. Remember that the
Newton’s third law of motion
Newton’s third law of motion is about interacting objects. These could be two magnets attracting or repelling each other, two electrons repelling each other, etc. Newton’s third law states:
When two bodies interact, the forces they exert on each other are equal and opposite.
How can we relate this to the idea of momentum? Picture holding two magnets, one in each hand. You gradually bring them towards each other (Figure 6.21) so that they start to attract each other. Each feels a force pulling it towards the other. The two forces are the same size, even if one magnet is stronger than the other. Indeed, one magnet could be replaced by an unmagnetised piece of steel and they would still attract each other equally.
If you release the magnets, they will gain momentum as they are pulled towards each other. One gains momentum to the left while the other gains equal momentum to the right.
Each is acted on by the same force, and for the same time. Hence momentum is conserved.
B
force of A on B
A force of B on A
Figure6.21 Newton’sthirdlawstatesthattheforcesthesetwo
magnets exert one each other must be equal and opposite.
15 Water pouring from a broken pipe lands on a flat roof. The water is moving at 5.0 m s−1 when it strikes the roof. The water hits the roof at a rate of 10 kg s−1. Calculate the force of the water hitting the roof. (Assume that the water does not bounce as it hits the roof. If it did bounce, would your answer be greater or smaller?)
16 A golf ball has a mass of 0.046 kg. The final velocity of the ball after being struck by a golf club is 50 m s−1. The golf club is in contact with the ball for a time of 1.3 ms. Calculate the average force exerted by the golf club on the ball.
Chapter 6: Momentum
equation F = ma is a special case of F = Δp which only Δt
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applies when the mass of the object is constant. There are situations where the mass of an object changes as it moves, for example a rocket, which burns a phenomenal amount of chemical fuel as it accelerates upwards.
QUESTIONS
13 A car of mass 1000 kg is travelling at a velocity of
10 m s−1. It accelerates for 15 s, reaching a velocity of 24 m s−1. Calculate:
a the change in the momentum of the car in the 15 s period
b the average force acting on the car as it accelerates.
14 A ball is kicked by a footballer. The average force on the ball is 240 N and the impact lasts for a time interval of 0.25 s.
a Calculate the change in the ball’s momentum.
b State the direction of the change in momentum.
N
S