A rod of mass M and length L is lying on a horizontal frictionless surface. A particle of mass m travelling along the surface hits at one end of the rod with velocity u in a direction perpendicular to the rod. The collision is completely elastic. After collision, particle comes to rest. The ratio of masses (m/M) is 1/x . The value of x will be ____________.Correct answer is '4'. Can you explain this answer?

Question

Answer

Problem: Find the value of x where a particle of mass m collides with a rod of mass M and length L lying on a horizontal frictionless surface with velocity u in a direction perpendicular to the rod. The collision is completely elastic and after the collision, the particle comes to rest. The ratio of masses (m/M) is 1/x.

Solution:

Step 1: Analyzing the initial state

Before the collision, the particle of mass m is moving with a velocity u in a direction perpendicular to the rod. The rod of mass M and length L is lying on a horizontal frictionless surface at rest.

Step 2: Analyzing the collision

When the particle collides with the rod, it exerts a force on the rod. As the collision is completely elastic, the kinetic energy of the system is conserved. Therefore, the particle rebounds with the same velocity u in the opposite direction. By conservation of momentum, the rod acquires a velocity v in the same direction as that of the particle.

Step 3: Analyzing the final state

After the collision, the particle comes to rest and the rod is moving with a velocity v.

Step 4: Applying conservation of momentum

Before the collision, the momentum of the system was zero. After the collision, the momentum of the system is:

mv - M(0) = (m+M)v'

where v' is the velocity of the rod and particle after the collision.

Solving for v', we get:

v' = mv / (m+M)

Step 5: Applying conservation of energy

Before the collision, the kinetic energy of the system was:

(1/2)mu^2

After the collision, the kinetic energy of the system is:

(1/2)mv'^2 + (1/2)Mv'^2

Since the collision is completely elastic, the kinetic energy of the system is conserved. Therefore, we have:

(1/2)mu^2 = (1/2)(m+M)v'^2

Substituting the value of v', we get:

(1/2)mu^2 = (1/2)m^2v^2 / (m+M)^2 + (1/2)M^2v^2 / (m+M)^2

Simplifying, we get:

u^2 = v^2 / (m+M) + M^2v^2 / (m+M)^2

Step 6: Solving for x

Substituting the value of v' in terms of v, we get:

v^2 = 2mu^2 / (m+M)

Substituting this value in the previous equation, we get:

u^2 = 2mv^2 / (m+M) + 2Mv^2 / (m+M)

Simplifying, we get:

(m/M) = 3

Therefore, the value of x is:

x = 1 / (m/M) = 4

Hence, the value of x is 4.

Answer

Just before collision

Just after collision
From momentum conservation,

mu = Mv

From angular momentum conservation about O,

Coefficient of restitution = e = Relative velocity after collsion / Relative velocity before collision

From equations (ii) and (iii), we get

x = 4

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