(^616) „„„„„ A Textbook of Engineering Mechanics
Fig. 30.9.
From the geometry of the figure, we find that when the body swings
through 30° i.e. from A to B, it has gone up by a distance,
h = 0·8 – 0·8 cos 30°
= 0·8 – (0·8 × 0·866) m
= 0·1072 m.
We know that Kinetic energy of the body and bullet after
impact at A.
( )^22 (0·03 10)
N-m
22
mMv++×v

= 5·015 v^2 N-m ...(i)
and potential energy of the body at B
= (m + M) gh = (10 + 0·03) 9·8 × 0·1072 N-m
= 10·54 N-m ...(ii)
Since entire kinetic energy of the body and bullet is used in raising the body (from A to B),
therefore equating equations (i) and (ii),
5·015 v^2 = 10·54 or v = 1·45 m/s
We also know that momentum of the body and bullet just after impact
= (10 + 0·03) 1·45 = 14·54 kg-m/s ...(iii)
and momentum of the bullet just before impact
= 0·03 u kg-m/s ...(iv)
Now equating equations (iii) and (iv),
14·54 = 0·03 u
∴
14·54
484·7 m/s
0·03
u== Ans.
30.20. TRANSFORMATION OF ENERGY
In the previous articles we have discussed potential energy and kinetic energy. Now we shall
discuss the transformation of potential energy into kinetic energy. Consider a body just dropped on
the ground from A as shown in Fig. 30.10. Let us consider the ground level as the datum or reference
level.
Let m = Mass of the body, and
h = Height from which the body is
dropped.
Now consider other positions B and C of the same body at various
times of the fall. Now we shall find total energy of the body at these
positions.
Energy at A
Since the body at A has no velocity, therefore kinetic energy at A
= 0
and potential energy at A = mgh
∴Total energy at A = mgh
Energy at B
We know that at B, the body has fallen through a distance (y).
Therefore velocity of the body at B
= 2 gy
Fig. 30.10.