Mechanical Engineering Principles

LINEAR MOMENTUM AND IMPULSE 141

Massm=40 g= 0 .040 kg, initial velocity,

u=15 m/s and final velocity,v=−5 m/s (nega-

tive since the rebound is in the opposite direction to

velocityu)

(a) Momentum before impact

=mu= 0. 040 × 15 = 0 .6kgm/s

Momentum after impact

=mv= 0. 040 ×− 5 =− 0 .2kgm/s

Impulse=change of momentum

= 0. 6 −(− 0. 2 )= 0 .8kgm/s

(b) Impulsive force=

change of momentum

change of time

=

0 .8kgm/s

0. 20 × 10 −^3 s

=4000 Nor4kN

Problem 13. A gun of mass 1.5 t fires a

shell of mass 15 kg horizontally with a

velocity of 500 m/s. Determine (a) the initial

velocity of recoil, and (b) the uniform force

necessary to stop the recoil of the gun in

200 mm.

Mass of gun,mg= 1 .5t=1500 kg, mass of shell,

ms=15 kg, initial velocity of shell,us=500 m/s.

(a) Momentum of shell=msus= 15 × 500

=7500 kg m/s.

Momentum of gun=mgv=1500 v

wherev=initial velocity of recoil of the gun.

By the principle of conservation of momentum,

initial momentum=final momentum, i.e.

0 = 7500 +1500 v, from which,

velocityv=

− 7500

1500

=−5m/s

(the negative sign indicating recoil velocity)

i.e. the initial velocity of recoil=5m/s.

(b) The retardation of the recoil,a, may be deter-
mined usingv^2 =u^2 + 2 as,wherev,thefinal

velocity, is zero,u, the initial velocity, is 5 m/s

ands, the distance, is 200 mm, i.e. 0.2 m.

Rearrangingv^2 =u^2 + 2 asforagives:

a=

v^2 −u^2

2 s

=

02 − 52

2 ( 0. 2 )

=

− 25

0. 4

=− 62 .5m/s^2

Force necessary to stop recoil in 200 mm

=mass×acceleration

=1500 kg× 62 .5m/s^2

=93750 Nor 93 .75 kN

Problem 14. A vertical pile of mass 100 kg

is driven 200 mm into the ground by the

blow of a 1 t hammer which falls through

750 mm. Determine (a) the velocity of the

hammer just before impact, (b) the velocity

immediately after impact (assuming the

hammer does not bounce), and (c) the

resistive force of the ground assuming it to

be uniform.

(a) For the hammer,v^2 =u^2 + 2 gs,where

v=final velocity,u=initial velocity=0,

g= 9 .81 m/s^2 ands=distance=750 mm=

0 .75 m.

Hencev^2 = 02 + 2 ( 9. 81 )( 0. 75 ), from which,

velocity of hammer, just before impact,

v=

√

2 ( 9. 81 )( 0. 75 )= 3 .84 m/s

(b) Momentum of hammer just before impact

=mass×velocity

=1000 kg× 3 .84 m/s=3840 kg m/s

Momentum of hammer and pile after impact=

momentum of hammer before impact.

Hence, 3840 kg m/s=(mass of hammer and

pile)×(velocity immediately after impact)

i.e. 3840=( 1000 + 100 )(v),from which,

velocity immediately after impact ,

v=

3840

1100

= 3 .49 m/s.