Mechanical Engineering Principles

LINEAR MOMENTUM AND IMPULSE 137

30 m/s= 30 × 3 .6km/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

Massm 1 =10 t=10000 kg,m 2 =15000 kg and
velocityu 1 =6m/s,u 2 = 0.

Total momentum before impact

=m 1 u 1 +m 2 u 2

=( 10000 × 6 )+( 15000 × 0 )=60000 kg m/s

Let the common velocity of the wagons after impact
bevm/s
Since total momentum before impact = total
momentum after impact:

60000 =m 1 v+m 2 v

=v(m 1 +m 2 )=v( 25000 )

Hence v=

60000

25000

= 2 .4m/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.

Massm 1 =30 g= 0 .030 kg,

m 2 =20 g= 0 .020 kg,velocityu 1 =20 m/s and

u 1 =15 m/s.

(a) When the velocities have the same line of

action and the same sense, bothu 1 andu 2 are

considered as positive values

Total momentum before impact

=m 1 u 1 +m 2 u 2

=( 0. 030 × 20 )+( 0. 020 × 15 )

= 0. 60 + 0. 30 = 0 .90 kg m/s

Let the common velocity after impact bevm/s

Total momentum before impact = total

momentum after impact

i.e. 0. 90 =m 1 v+m 2 v=v(m 1 +m 2 )

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 massm 1 as positive gives:

velocityu 1 =+20 m/s andu 2 =−15 m/s

Total momentum before impact

=m 1 u 1 +m 2 u 2

=( 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 massm 1 .If

the common velocity after impact isvm/s then

0. 30 =v(m 1 +m 2 )=v( 0. 050 )