86
2 Particle Dynamics
Eliminating p H between (1) and (2) and simplifying
P L =
2 p 0 m
M + m
mv L =
2Mum
M + m
or v L =
2u M
M + m
= 2u (∵ m << M)
2.40 Let a body of mass m 1 moving with velocity u make a completely inelastic
collision with the body of mass m 2 initially at rest. Let the combined mass
moves with a velocity v c given by
v c =
m 1 u
m 1 + m 2
=
u
2
(∵ m 1 = m 2 )
Energy lost
=
1
2
mu
2
−
1
2
(2m)
u
2
2 =
1
4
mu
2
=
1
2
K 0
where K 0 =
1
2 mu 2 is the initial kinetic energy.
2.41 Let the speed of the bullet be u. Let the block + bullet system be travelling
with initial speed v. If m and M are the masses of the bullet and the block,
respectively, then momentum conservation gives
mu = (M + m)v
(1)
∴ v =
mu
M + m
(2)
The initial kinetic energy of the block + bullet system
K =
1
2
(M + m)v
2
=
1
2
m 2 u 2
(M + m)
Work done to bring the block + bullet system to rest in distance s is
W = μ(M + m)gs =
1
2
m 2 u 2
(M + m)
∴ u =
(M + m)
m
2μ gs =
(2.000 + 0.005)
0.005
√
2 × 0.2 × 9.8 × 2
= 1123m/s
2.42 (a) Let m 1 = m with velocity u collide with m 2 = M, initially at rest. For
elastic collision the final velocities will be
v 1 =
(m 1 − m 2 )
m 1 + m 2
u =
(m − M)
m + M
u
(m < M)
(1)