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A uniform thin rod of mass 3 kg and length 2 m lies at rest on a…

Question

A uniform thin rod of mass 3 kg and length 2 m lies at rest on a frictionless horizontal table. A point particle of mass 1 kg, moving on the table with speed 8 m/s in a direction perpendicular to the rod, strikes one end of the rod and sticks to it. After the collision the combined body rotates with angular speed equal to p/q rad/s in lowest terms, where p and q are positive integers. Find p + q.

✓ Verified answer: 31checked by our engine — not a guess

Step-by-step solution

After sticking, momentum and angular momentum are conserved (no external horizontal force/torque).

Locate the combined centre of mass: from the rod's centre it is d = m(L/2)/(M+m) = 1*(1)/4 = 0.25 m toward the impact end.

Distances from the system CM: particle r_p = L/2 - d = 1 - 0.25 = 0.75 m; rod centre r_rod = 0.25 m.
Moment of inertia about the system CM: I = [M L^2/12 + M r_rod^2] + m r_p^2 = [3*4/12 + 3*0.0625] + 1*0.5625 = [1 + 0.1875] + 0.5625 = 1.75 kg m^2 = 7/4.
Angular momentum about the system CM before impact = m v r_p = 1*8*0.75 = 6 kg m^2/s.
So omega = 6 / (7/4) = 24/7 rad/s.
Thus p/q = 24/7 and p + q = 31.

Final answer31

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