A thin uniform rod of mass m and length l is kept on a smooth horizontal surface such that it can move freely. At what distance from centre of rod should a particle of mass m strike on the rod such that the point P at a distance l/3 from the end of the rod is instantaneously at rest just after the elastic collision?

Complete Solution:

A thin uniform rod of mass m and length l is placed on a smooth horizontal surface. A particle of mass m strikes the rod at some distance x from its center. After the collision, the point P, located at l/3 from one end of the rod (or l/6 from its center), must be instantaneously at rest.

Conservation of Linear Momentum:

Total momentum before and after the collision must be conserved:  mv₀ = mv + MV_cm

Here:

• m = mass of the particle
• v₀ = initial velocity of the particle
• v = final velocity of the particle
• M = m, mass of the rod
• V_cm = velocity of the rod's center of mass after the collision

Conservation of Angular Momentum:

Angular momentum about the center of the rod before and after the collision must also be conserved: mv₀ x = mv x + (1/12)M l² ω

where ω is the angular velocity of the rod after the collision.

Condition for Point P:

The velocity of point P, located at l/6 from the center of the rod, is: V_P = V_cm - ω (l/6)

For P to be instantaneously at rest: V_P = 0 implies V_cm = ω (l/6)

Using the above conditions:

From conservation of linear momentum: mv₀ = mv + mV_cm

From conservation of angular momentum: mv₀ x = mv x + (1/12)m l² ω

From the condition for P to be at rest: V_cm = ω (l/6)

Substitute ω = 6V_cm/l into the angular momentum equation and solve. After simplifying, the distance x at which the particle must strike is: x = l/2

Final Answer:

The particle must strike the rod at a distance l/2 from the center for point P to be at rest.