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# A Uniform Rod of Length L Lies on a Smooth Horizontal Table. a Particle Moving on the Table Strikes the Rod Perpendicularly at an End and Stops. - Physics

Sum

A uniform rod of length L lies on a smooth horizontal table. A particle moving on the table strikes the rod perpendicularly at an end and stops. Find the distance travelled by the centre of the rod by the time it turns through a right angle. Show that if the mass of the rod is four times that of the particle, the collision is elastic.

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#### Solution

Let initial velocity of the particle = u1

Let final velocity of the particle = v1

Let time taken to move π/2 angle = t

Given v1 = 0

Let initial velocity of CM of the rod =  u2

Given u2 = 0

Let final velocity of the CM of rod = v2

By conservation of linear momentum

mu1 + Mu2 = mv1 + Mv2

=> v_2 = (m/M)u_1 ...........(1)

Now by we consider the angular momentum imparted by the particle to the rod.

m u_1(L/2)  = Iω

For rod about its CM, I = (ML_2)/12

mu1(L/2)  = ωML2/12

=> ω = (6m u_1)/(ML)  .........(2)

Now we know

ω = θ/t

θ = π/2

=> t = θ/ω = (π/2)/((6m u_1)/(ML))

=> t = (πML)/(12m u_1)

Linear distance moved by the CM of the rod will be
s = v2t

By equation 1.

s = [(πML)/(12m u_1)]× (m/M)u_1

= (πL)/12

Initial KE of the particle is ½m(u1)2

"KE of the CM of the rod "=  1/2M(v_1)^2 = 1/2M{(m/M)u_1}^2  ......["by equation 1"]

= 1/2(u_1)^2m_2/M

"KE"_"rot" " of the rod "= 1/2Iω^2

"KE"_"rot" = 1/2 ((ML_2)/12)((6m u_1)/(ML))^2

=  (3/2) m^2(u_1)^2/M

To be elastic collision KE must be conserved

1/2m(u_1)^2 = 1/2(u_1)^2m^2/M + (3/2) m^2(u_1)^2/M

=> M = 4m

Hence Proved.

Concept: Centre of Mass of a Uniform Rod
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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 10 Rotational Mechanics
Q 60 | Page 199
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