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4.1Newtonâs second law in terms of momentum
In ChapterÂ 2, Newtonâs second law was expressed as FâÂ =maâÂ . In fact, this equation is only valid in those cases where the mass of the system remains constant. But there are plenty of situations where the mass changes, and in those situations, we must use a different version of the second law. Example situations include:
âthe motion of a rocket where the mass decreases due to burnt fuel ejected away from the rocket
âsand falling on a conveyor belt
âa droplet of water falling through mist and increasing in mass.
We define linear momentum, pâÂ , to be the product of the mass of a body and its velocity:
pâÂ =mvâÂ 
Momentum is a vector; note that the direction of the momentum is the direction of the velocity.
Its unit is kg m sâ1. Because kg m sâ1 = kg m sâ2 s = N s, the unit of momentum is also N s.
In terms of momentum, Newtonâs second law is FâÂ net=ÎpâÂ Ît: the net force on a system is equal to the rate of change of the systemâs momentum.
We can see that, if the mass stays constant, then this version reduces to the usual maâÂ :
FâÂ net=ÎpâÂ Ît=pâÂ FâpâÂ IÎt=mvâÂ FâmvâÂ IÎt=m(vâÂ FâvâÂ IÎt)=mÎvâÂ Ît=maâÂ 
CHECK YOURSELF 1
A body of mass 4.0 kg moving horizontally at 8.0 m sâ1 bounces off a vertical wall with speed 3.0 m sâ1.
What is the magnitude of the change in the bodyâs momentum?
WORKED EXAMPLE 4.1
This is not an easy worked example, but it is important to remember that you must be careful when the mass changes.
A cart moves in a horizontal line with constant speed v. Rain starts to fall and the cart fills with water at a rate of Ï kg sâ1. (This means that in 1 second, Ï kg have fallen on the cart.) The cart must keep moving at constant speed. Determine the force that must be applied on the cart.
Answer
Notice that, if Fnet = ma (we drop the arrows of the vector notation) were valid, the force would have to be zero since there is no acceleration. But we need a force to act on the cart because the momentum of the cart is increasing (because the mass is).
The force is Fnet=ÎpÎt=Î(mv)Ît=vÎmÎt=vÏ.
So, if Ï = 0.20 kg sâ1 and v = 3.5 m sâ1, the force would have to be 0.70 N.
EXAM TIP
You must be careful when mass changes. Zero acceleration does not imply zero net force in this case.
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WORKED EXAMPLE 4.2
A 0.50 kg ball bounces vertically off a hard surface. A graph of velocity against time is shown in Figure 4.1.

Figure 4.1: For worked example 4.2.

aFind the magnitude of the momentum change of the ball during the bounce.
bThe ball stayed in contact with floor for 0.15 s. What average force did the ball exert on the floor?
Answer
aThe initial momentum is 0.50 Ã 4.0 = 2.0 N s.
The final momentum is 0.50 Ã (â2.0) = â1.0 N s.
The magnitude of the change is, therefore, 3.0 N s.
bThe forces on the ball during contact are its weight and the normal force from the floor:
FnetÂ =Râmg=ÎpÎt=3.0Ns0.15Â s=20Â N
So, R = 20 + 5.0 = 25 N.
EXAM TIP
This is a tricky problem with lots of possibilities for error. Many students forget to include the minus sign in the rebound velocity and/or forget the weight and answer incorrectly that R = 20 N.
TEST YOUR UNDERSTANDING
1The momentum of a ball increased by 12.0 N s as a result of a force that acted on the ball for 2.00 s. Find the average force on the ball.
2A ball of mass 2.0 kg rolling on a horizontal floor with a speed 4.0 m sâ1 hits a vertical wall. Calculate the magnitude of the change in momentum of the ball if it
abounces with speed 4.0 m sâ1
bbounces with speed 3.0 m sâ1
csticks to the wall.
3A ball of mass m is dropped vertically from a height h1 and rebounds to a height h2. The ball is in contact with the floor for a time interval of Ï.
aShow that the average net force on the ball is given by
F=m2gh1âââââ+2gh2âââââÏ
bIf h1 = 8.0 m, h2 = 6.0 m, Ï = 0.125 s and m = 0.250 kg, calculate the average force exerted by the ball on the floor.
4Water exits a horizontal pipe at a rate of 40 kg per minute with speed 9.0 m sâ1.
What horizontal force is needed to keep the pipe in place?
