(^562) A Textbook of Engineering Mechanics
Loss of kinetic energy during impact
We know that loss of kinetic energy during impact,
12 22
12
12
(– )(1– )
L 2( )
mm
Euue
mm
(^15) (3 – 0.6) (1 – 0.75 ) N-m 22
2(1 5)
×
(^5) (2.4) (^2) 0.4375 1.05 kg-m
12
=×== 1.05 J Ans.
Final velocity of the first sphere0
Let v 1 = Final velocity of the first sphere, and
v 2 = Final velocity of the second sphere.
We know from the law of conservation of momentum that
m 1 u 1 + m 2 u 2 = m 1 v 1 + m 2 v 2
(1 × 3) + (5 × 0.6) = 1 × v 1 + 5 × v 2
∴ v 1 + 5v 2 = 6 ...(i)
We also know from the law of collision of elastic bodies that
(v 2 – v 1 ) = e (u 1 – u 2 ) = 0.75 (3 – 0.6) = 0.75 × 2.4
or v 2 – v 1 = 1.8 ...(ii)
Adding equations (i) and (ii),
6 v 2 = 7.8
∴ 2
7.8
1.3 m/s
6
v ==
Substituting this value of v 2 in equation (i),
v 1 + (5 × 1.3) = 6
or v 1 = 6 – (5 × 1.3) = – 0.5 m/s
Minus sign indicates, that direction of motion of the first body is reversed after impact. Ans.
EXERCISE 27.1
- A ball of mass 2 kg impinges directly with a ball of mass 1 kg, which is at rest. If the
velocity of the smaller mass after impact, be the same as that of the first ball before
impact, find the coefficient of restitution. (Ans. 0.5) - Two balls of masses 2 kg and 3 kg are moving with velocities 2 m/s and 3 m/s towards
each other. If the coefficient of restitution is 0.5, find the velocity of the two balls after
impact. (Ans. 2.5 m/s ; 0) - Three spheres of masses 2 kg, 6 kg and 12 kg are moving with velocities of 12 m/s, 4 m/s
and 2 m/s respectively in a straight line. Show that after impact of first and second as well
as second and third, the first two spheres will be brought to rest. Take e = 1.0. - A bullet of mass 50 gm is fired into a freely suspended target of 2.5 kg. On impact, the
target moves with a velocity of 2.5 m/s. Find the velocity of the bullet and the loss of
kinetic energy, if the impact is perfectly inelastic. (Ans. 127.5 m/s ; 398.4 kg-m)
Hint. After impact, the bullet and target will move with the same velocity.