Linear momentum and impulse At the end of this chapter you should be
12.1 Linear momentum
Themomentumof a body is defined as the product of its mass and its velocity, i.e.momentum=mu, wherem=mass (in kg) andu=velocity (in m/s). The unit of momentum is kg m/s
Since velocity is a vector quantity, momentum is a vector quantity, i.e. it has both magnitude and direction.
Newton’s first law of motionstates:
a body continues in a state of rest or in a state of uniform motion in a straight line unless acted on by some external force
Hence the momentum of a body remains the same provided no external forces act on it.
The principle of conservation of momentumfor a closed system (i.e. one on which no external forces act) may be stated as:
the total linear momentum of a system is a constant The total momentum of a system before collision in a given direction is equal to the total momentum
m2 m1 u1 u2 Mass Mass Figure 12.1
of the system after collision in the same direction. In Figure 12.1, masses m1 and m2 are travelling in the same direction with velocity u1 > u2. A collision will occur, and applying the principle of conservation of momentum:
total momentum before impact
=total momentum after impact i.e. m1u1+m2u2=m1v1+m2v2
where v1 and v2 are the velocities of m1 and m2 after impact.
Problem 1. Determine the momentum of a pile driver of mass 400 kg when it is moving downwards with a speed of 12 m/s.
Momentum=mass×velocity =400 kg×12 m/s
=4800 kg m/s downwards Problem 2. A cricket ball of mass 150 g has a momentum of 4.5 kg m/s. Determine the velocity of the ball in km/h.
Momentum=mass×velocity, hence velocity= momentum
mass = 4.5 kg m/s
30 m/s=30×3.6 km/h
=108 km/h=velocity of cricket ball.
Problem 3. Determine the momentum of a railway wagon of mass 50 tonnes moving at a velocity of 72 km/h.
Momentum=mass×velocity
Mass=50 t=50000 kg (since 1 t=1000 kg) and velocity=72 km/h= 72
3.6 m/s=20 m/s. Hence,momentum=50000 kg×20 m/s
=1000000 kg m/s =106 kg m/s Problem 4. A wagon of mass 10 t is moving at a speed of 6 m/s and collides with another wagon of mass 15 t, which is stationary. After impact, the wagons are coupled together. Determine the common velocity of the wagons after impact
Massm1 = 10 t = 10000 kg, m2 = 15000 kg and velocityu1=6 m/s, u2 = 0.
Total momentum before impact =m1u1+m2u2
=(10000×6)+(15000×0)=60000 kg m/s
Let the common velocity of the wagons after impact bev m/s
Since total momentum before impact = total momentum after impact:
60000=m1v+m2v
=v(m1+m2)=v(25000)
Hence v = 60000
25000 =2.4 m/s
i.e.the common velocity after impact is 2.4 m/s in the direction in which the 10 t wagon is initially travelling.
Problem 5. A body has a mass of 30 g and is moving with a velocity of 20 m/s. It collides with a second body which has a mass of 20 g and which is moving with a velocity of 15 m/s. Assuming that the bodies both have the same velocity after impact, determine this common velocity, (a) when the initial velocities have the same line of action and the same sense, and (b) when the initial velocities have the same line of action but are opposite in sense.
Massm1 =30 g=0.030 kg,
m2 = 20 g = 0.020 kg,velocity u1 = 20 m/s and u1 =15 m/s.
(a) When the velocities have the same line of action and the same sense, bothu1 andu2 are considered as positive values
Total momentum before impact =m1u1+m2u2
=(0.030×20)+(0.020×15)
=0.60+0.30=0.90 kg m/s
Let the common velocity after impact bev m/s Total momentum before impact = total momentum after impact
i.e. 0.90=m1v+m2v=v(m1+m2) 0.90=v(0.030+0.020)
from which, common velocity, v = 0.90 0.050 = 18 m/s in the direction in which the bodies are initially travelling
(b) When the velocities have the same line of action but are opposite in sense, one is consid- ered as positive and the other negative. Taking the direction of mass m1 as positive gives: velocity u1 = +20 m/s andu2= −15 m/s Total momentum before impact
=m1u1+m2u2
=(0.030×20)+(0.020× −15)
=0.60−0.30= +0.30 kg m/s
and since it is positive this indicates a momen- tum in the same direction as that of massm1. If the common velocity after impact isv m/s then
from which, common velocity,v= 0.30 0.050 = 6 m/s in the direction that the 30 g mass is initially travelling.
Problem 6. A ball of mass 50 g is moving with a velocity of 4 m/s when it strikes a stationary ball of mass 25 g. The velocity of the 50 g ball after impact is 2.5 m/s in the same direction as before impact. Determine the velocity of the 25 g ball after impact. Massm1 =50 g=0.050 kg,
m2=25 g=0.025 kg. Initial velocityu1=4 m/s, u2 =0; final velocityv1=2.5 m/s,v2 is unknown. Total momentum before impact
=m1u1+m2u2
=(0.050×4)+(0.025×0)
=0.20 kg m/s
Total momentum after impact =m1v1+m2v2
=(0.050×2.5)+(0.025v2)
=0.125+0.025v2
Total momentum before impact=total momentum after impact, hence
0.20=0.125+0.025v2
from which,velocity of 25 g ball after impact, v2=
0.20−0.125
0.025 =3 m/s
Problem 7. Three masses,P,Qand Rlie in a straight line.P has a mass of 5 kg and is moving towardsQat 8 m/s.Qhas a mass of 7 kg and a velocity of 4 m/s, and is moving towards R. Mass Ris stationary.P collides withQ, and P andQthen collide with R. Determine the mass ofR assuming all three masses have a common velocity of 2 m/s after the collision ofP andQ with R.
MassmP =5 kg,mQ=7 kg, velocityuP =8 m/s anduQ =4 m/s.
Total momentum beforeP collides with Q=mPuP+mQuQ
=(5×8)+(7×4)=68 kg m/s
Let P and Q have a common velocity of v1 m/s after impact.
Total momentum afterP and Qcollide =mPv1+mQv1
=v1(mP+mQ)=12v1
Total momentum before impact=total momentum after impact, i.e. 68= 12v1, from which, common velocity ofP and Q, v1 = 68 12 =5 2 3 m/s.
Total momentum afterP and Qcollide withR
=(mP+Q×2)+(mR×2)(since the common velocity after impact=2 m/s)
=(12×2)+(2mR)
Total momentum beforeP andQcollide withR= total momentum afterP and Qcollide withR, i.e. mP+Q×523
=(12×2)+2mR i.e. 12×523 =24+2mR
68−24=2mR from which, mass of R, mR= 44
2 =22 kg. Now try the following exercise
Exercise 59 Further problems on linear