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4.4Kinetic energy and momentum

We have seen that, in a collision or explosion where no external forces are present, the total momentum of the system is conserved. You can easily convince yourself that in the three collisions illustrated in Figure 4.15 momentum is conserved. The incoming body has mass 8.0 kg and the other a mass of 12 kg.

Figure 4.15: Momentum is conserved in these three collisions.

Let us examine these collisions from the point of view of energy.

In all cases the total kinetic energy before the collision is

E K=1 2×8×10 2=400 J

The total kinetic energy after the collision in each case is

case 1 E K=1 2×20×4 2=160 J

case 2 E K=1 2×8×1 2+1 2×12×6 2=220 J

case 3 E K=1 2×8×2 2+1 2×12×8 2=400 J

We observe that whereas momentum is conserved in all cases, kinetic energy is not.

●When kinetic energy is conserved (case 3), the collision is said to be an elastic collision a collision in which the total kinetic energy before the collision equals the total kinetic energy after the collision; that is, E K is conserved.

●When kinetic energy is not conserved (cases 1 and 2), the collision is said to be an inelastic collision a collision in which the total kinetic energy after a collision is less than the total kinetic energy before the collision; that is, some E K is lost during the collision.

In an inelastic collision, kinetic energy is ‘lost’. When the bodies stick together after a collision (case 1), the collision is said to be totally inelastic (or plastic) and, in this case, the maximum possible kinetic energy is ‘lost’. (In an explosion the kinetic energy afterward is greater than the initial kinetic energy.)

The ‘lost’ kinetic energy is transformed into other forms of energy, such as thermal energy, deformation energy (if the bodies are permanently deformed as a result of the collision) and sound energy.

Notice that using momentum we can obtain a useful additional formula for kinetic energy:

E K=1 2 m v 2=m 2 v 2 2 m=p 2 2 m

Check yourself 4

You are on a frictionless frozen lake and throw your shoes away in the same direction. Are the momentum and kinetic energy of the system consisting of you and the shoes conserved?

Worked example 4.5

A moving body of mass m collides with a stationary body of double the mass and sticks to it. Calculate the fraction of the original kinetic energy that is lost.

Answer

The original kinetic energy is 1 2 m v 2, where v is the speed of the incoming mass. After the collision, the two bodies move as one with speed u that can be found from momentum conservation:

m v=(m+2 m)u⇒u=v 3

So, the total kinetic energy after the collision is

1 2(3 m)(v 3)2=m v 2 6

So, the lost kinetic energy is

m v 2 2−m v 2 6=m v 2 3

And the fraction of the original energy that is lost is

(m v 2 3)(m v 2 2)=2 3

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Worked Example 4.6

A body at rest of mass M explodes into two pieces of masses m 4 and 3 m 4.

Calculate the ratio of the kinetic energies of the two fragments.

Answer

It helps to use the formula for kinetic energy in terms of momentum: E K=p 2 2 m.

The total momentum before the explosion is zero, so total momentum is also zero after the collision. Thus, the two fragments must have equal and opposite momenta. Hence,

E light E bexy =(p 2 2 M lipht )((−p)2 2 M leam )=M bany M light =(3 M 4)(M 4)=3

In an explosion, the kinetic energy increases. The increase in energy comes from the chemical energy of the explosives.

Worked example 4.7

This popular exam question requires the same calculations as worked example 4.1, but it is worth doing again. Gravel falls vertically on a conveyor belt at a rate of σ kg s–1, as shown in Figure 4.16.

Figure 4.16: For worked example 4.7.

aDetermine:

ithe force that must be applied on the belt to keep it moving at constant speed v

iithe power that must be supplied by the motor turning the belt

iiiThe rate at which the kinetic energy of the gravel is changing.

bExplain why the answers to parts ii and iii are different.

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Answer

a iThe force is F net=Δ p Δ t=Δ(m v)Δ t=v Δ m Δ t=v σ.

iiThe power is found from P = F v to be P = (σ v) v = σ v 2.

iiiIn 1 second the mass on the belt increases by σ kg. The kinetic energy of this mass is E K=1 2 σ v 2. This is the increase in kinetic energy in a time of 1 s so the rate of kinetic energy increase is 1 2 σ v 2.

bThere must be a frictional force between the gravel and the belt. The gravel slides on the belt for some time until it picks up the speed of the belt. During this sliding heat is being generated. So, of the power σ v 2 that is being supplied by the motor driving the belt, half goes into increasing the kinetic energy (due to the increasing mass) and the other half is wasted as heat.

More precisely, let an amount of mass Δm fall on the belt. A frictional force f will provide an acceleration a=f Δ m to the mass so that over a distance d the mass will reach the speed of the belt. Thus v 2=2 a d=2 f Δ m d. This says that the work done by the frictional force is f d=1 2 Δ m v 2, and so the rate at which friction dissipates energy is 1 2 Δ m Δ t v 2=1 2 σ v 2.

Check yourself 5

Water leaves a horizontal pipe of cross-sectional area 5 × 10−3 m2 at a rate of 20 kg s–1 and speed 15 m s–1. What force keeps the pipe fixed?

TEST YOUR UNDERSTANDING

12In a collision of two blocks, the blocks exert forces on each other. Explain why the momentum of the system consisting of the two blocks is conserved in the collision.

13Two masses of 2.0 kg and 4.0 kg are held in place, compressing a spring between them. When they are released, the 2.0 kg mass moves away with a speed of 3.0 m s−1. What was the energy stored in the spring?

14Two masses m and M are held in place, compressing a spring between them. When they are released, m moves to the left with speed v. What is the change in momentum of M?

15The bodies in Figure 4.17 suffer a head-on collision and stick to each other afterwards. Find their common velocity.

Figure 4.17: For question 14.

16Two masses moving in a straight line toward each other collide as shown in Figure 4.18. Find the velocity (magnitude and direction) of the heavier mass after the collision.

Figure 4.18: For question 15.

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17A mass of 6.0 kg moving at 4.0 m s–1 collides with a stationary mass of 8.0 kg and sticks to it. How much kinetic energy was ‘lost’ in the collision?

18A mass of 6.0 kg moving at 14 m s–1 collides head on with a stationary mass of 8.0 kg. The lighter mass rebounds with a speed of 2.0 m s–1. Is this collision elastic?

19A body of mass M breaks apart at rest into two pieces of mass M/4 and 3M/4. The energy released in the break-up is Q. What is the kinetic energy carried by the lighter piece?

20A ball of mass 0.250 kg moving on a horizontal floor collides with a vertical wall as shown in Figure 4.19. The speed of the ball before and after the collision is 4.00 m s–1.

Calculate the magnitude and direction of the momentum change of the ball.

Figure 4.19: For question 19.

21A block of mass m slides from rest down a curved wedge of mass M (Figure 4.20). The wedge is on a frictionless table. The block is released from a height h.

Figure 4.20: For question 21.

When the block gets on the table its speed is v.

aDetermine, in terms of v, the speed with which the wedge recoils.

bShow that v 2=2 g h 1+m M.

22The rotor of a drone pushes air downward with speed v at a rate of 1.80 kg per second. The mass of the drone is 1.50 kg. The drone hovers stationary in air.

aDescribe the origin of the lift force on the drone.

bCalculate v while the drone hovers in air.

cCalculate the average power delivered to the air by the rotor.

dThe rotor is made to rotate faster, increasing the lift force by 50%. What is the acceleration of the drone?

23A bullet of mass 0.080 kg moving at speed v gets embedded in a block of mass 2.0 kg that hangs from a vertical string of length 1.5 m (Figure 4.21). The block and the bullet move together and stop when the string makes an angle of 34° with the vertical. What is v?

Figure 4.21: For question 23.