Question

Two particles, each of mass M and moving with speed y as shown. They
simultaneously strike the ends of a uniform rod of mass M and length d which is
pivoted at its center. The particles stick to the ends of the rod Find the ratio of total
initial kinetic energy of the two particles with the total loss in kinetic energy in the
collision of the two particles with the rod
M
0
d
Pivot
M
M​

Answer 1

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Answer 2

Answer:

The ratio of the total initial kinetic energy of the two particles with the total loss in kinetic energy in the collision of the two particles with the rod is 7:1.

Explanation:

CASE I: Before striking

The kinetic energy of particles before striking the rod will be in terms of the speed by which they are moving.

As per the question, the masses of the particles are M. Let's consider the speed at which they are moving as ' V'.

Now the initial kinetic energy (K₁) will be,

$K$₁$=1/2$×$MV^{2} +1/2$×$MV^{2} =MV^{2}$   (1)

CASE II: After striking

When the particles stick to the ends of the rod then they will start to act as a system rotating with angular speed (ω). Then the final kinetic energy (K₂)of the system will be,

$K$₂$=1/2$×$I$×$($ω$)^{2}$           (2)

Where,

I=moment of inertia of the system

ω=angular velocity

System=Two particles each of mass M+The Rod

As the net torque(τ) acting on the system is zero, the system's angular momentum is conserved.

So,

Initial angular momentum(L₁) = Final angular momentum(L₂)     (3)

$L$₁$=MV$$R+MVR=2MVR$     (4)

Here, R=distance between the hinged point i.e.center to the edge of the rod=d/2.

By putting R=d/2 in equation (4) we get;

$L$₁$=MVd$

$L$₂$=I$ω

Lets put the above values in equation (3);

$MVd=I$ω

ω$=MVd/I$       (5)

By putting equation (5) in equation (2) we get;

$K$₂$=M^{2}V^{2} d^{2} /2I$      (6)

Now we need to find 'I'.

Moment of inertia (I)=

Mass of the bodies×(distance of the body from the hinged point)²

$I=I$₁$+I$₂   (7)

$I$₁$=2$×$M(d/2)^{2}$  (for the two particles)

$I$₂$=Md^{2} /12$     (for the rod)

By substituting the above values in equation (7) we get;

$I=2$×$M(d/2)^{2} +Md^{2} /12$

$I=7/12$×$Md^{2}$     (8)

By putting equation (8) into equation (2) we get;

$K$₂$=6/7$×$MV^{2}$  (9)

Loss in kinetic energy(ΔK)=K₁-K₂    

ΔK=$MV^{2} -6/7$×$MV^{2}$$=MV^{2} /7$

Total initial kinetic energy(K₁)$/$Loss in kinetic energy(ΔK)

   $=MV^{2}$×$7/MV^{2}$$=7/1$

Hence, the ratio of the total initial kinetic energy of the two particles with the total loss in kinetic energy in the collision of the two particles with the rod is 7:1.