Three block
s are placed on smooth horizontal surface and lie on same horizontal straight line Block 1
and block 3 have mass m each and block 2 has mass M(M> m). Block 2 and block 3 ara initially
stationary, while block 1 is initially moving towards block 2 with speed v as shown. Assume that all
collisions are headon and perfectly elastic. What value of
ensures that block 1 and block 3 have the
same final speed ?
M
m
2
3
m
m
(A) 5.2
(B) 5-12
(C) 2.5
(D) 3.6
Particle "A moves with speed 10 m/s in a frictionless circular fixed horizontal pipe of radius 5 mang
Answers
Answer:
Initially collision occurs between block-1 and block-2
Let v
1
and v
2
are velocities after collision for block -1 and block-2 respectively.
Consider a=M/m.
Due to momentum conservation,
mv=mv
1
+Mv
2
or v=v
1
+av
2
...(1)
Due to energy conservation,
2
1
mv
2
=
2
1
mv
1
2
+
2
1
mv
2
2
+
2
1
Mv
2
2
or v
2
=v
1
+av
2
2
...(2)
By solving (1) and (2), we get
V
1
=
1+α
1−α
v ...(3)
v
2
=
1+α
2
v ...(4)
Now block-2 collides with block-3.
Let v
3
and v
4
are velocities after collisions of block-2 and block-3 respectively.
By Conservation of Momentum,
Mv
2
=Mv
3
+mv
4
or av
2
=av
3
+v
4
or a
4
a(v
2
−v
3
) ...(5)
By Conservation of energy,
2
1
Mv
2
2
=
2
1
Mv
3
2
+
2
1
mv
4
2
or av
2
2
=av
3
2
+v
4
2
....(6)
using eqns. (5) and (6), we get quadratic eqn.
(1+α)×v
3
2
−(2×α×v
2
)×v
3
+(α−1)×v
2
2
...(7)
Two solutions of the above quadratic equation are, v
2
=v
2
and v
2
=
α+1
α−1
v
2
the solution v
2
=v
2
is unacceptable becuase it gives v
4
=0. Hence if we consider the other solution, we get v
4
=
(1+a)
2
4a
Now if we equate the magnitude of v
4
and v
1
, we get
(1+α)2
4α
=
1+α
α−1
...(8)
[it is to be noted if a>1, sign of v
1
is negative, hence we copared the magnitude of v
1
and v
4
in eqn.(8)]
solving eqn.(8) for α, we get an aceeptable solution. a=2+
5
or (M/m)=2+
5