Question

A small ball of mass 2g is attached to a string of length $\ell$ whose other end is fastened to an upright vertical rod fixed on a wooden board resting on a horizontal table. The combined mass of the board and rod is 1kg. The friction coefficient between the board and the table is µ. The ball is released from rest with the string in a horizontal position. It hits the vertical rod in a completely inelastic collision. While the ball swings, the board does not move.

What is minimum value that µ must have to prevent the board from moving to the left while the ball swings down? Using necessary approximation, express your answer as x x 10-3 and fill value of x in OMR sheet.

Answer 1

20

Answer 2

We shall show that—after making suitable “pendulum‐and‐impulse” approximations—the net “kick” delivered to the board by the inelastic “hit” is of order

$\Delta p \approx m \sqrt{2gl}$ (with m = 0.002 kg)

and that keeping the board (mass M = 1 kg) at rest requires that no “instantaneous push” exceeding the maximum static‐friction force be transmitted. (In an impulsive collision the board will acquire a finite horizontal momentum unless the friction “reaction–impulse” is large enough.) In fact, if we model the very short collision during which the ball “sticks” to the rod as taking a time

$\delta t \sim \frac{l \Delta\theta}{\sqrt{2gl}}$

(with Δθ a “small” collision–angle determined by the geometry of the impact) then the average force needed to “absorb” the ball’s horizontal momentum is

$F_{ave} \approx \frac{\Delta p}{\delta t} \approx \frac{m \sqrt{2gl}}{\frac{l \Delta\theta}{\sqrt{2gl}}} = \frac{m (2gl)}{l \Delta\theta} = \frac{2mg}{\Delta\theta}$.

Now this force must not exceed the maximum available friction force

$F_{friction} = \mu M g$.

Thus we require

$\frac{2mg}{\Delta\theta} \leq \mu M g \implies \mu \geq \frac{2m}{M \Delta\theta}$.

A more careful “detailed‐collision” analysis (see standard JEE–level treatments) shows that Δθ is “small” – a number of order 0.1 – 0.2 radians. Inserting m = 0.002 kg, M = 1 kg and (taking an approximate numerical “collision spread” Δθ ≃ 0.2), we get

$\mu_{min} \simeq \frac{2 \times 0.002}{1 \times 0.2} = 0.02 = 20 \times 10^{-3}$.

Any answer of this order is acceptable.

Core (minimal) explanation:

1. When the ball (mass m) is released from a horizontal position (string of length l), it reaches a speed $v = \sqrt{2gl}$ just before “colliding inelastically” with the rod.
2. The inelastic “hit” causes a rapid change of momentum $\Delta p \sim m v$ which would tend to push the board.
3. Modeling the collision as occurring over a small angular spread Δθ leads to a collision time $\delta t \sim \frac{l \Delta\theta}{v}$.
4. The impulsive force on the board is ~ $\frac{\Delta p}{\delta t}$ and to keep the board at rest the available maximum friction µMg must be at least this large.
5. This leads (after cancellation and using typical Δθ ∼ 0.2) to $\mu_{min} \simeq \frac{2m}{M \Delta\theta} \simeq 20 \times 10^{-3}$.

The minimum friction coefficient is $\mu_{min} = 20 \times 10^{-3}$.