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Byju's Answer

Standard XII

Physics

Block on a Block Problems

A billiard ba...

Question

A billiard ball is struck by a cue. The line of action of applied impulse is horizontal and passes through the center of the ball. The initial velocity $V_{0}$ of the ball, radius $R$ , mass $m$ and coefficient of friction $μ$ between ball and table are known. The ball moves distance $\frac{12 V_{0}^{2}}{X μ g}$ before it ceases to slip on the table. What is the value of $X$ ?

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Solution

The correct option is A 49
$v_{0}^{2} - v^{2} = 2 a s = 2 μ g s v_{0} - v = μ g t v = R w T o r q u e = I n e r t i a \times α μ m g R = 0.4 m R^{2} α => α = 2.5 μ g / R w = α t R w = 2.5 μ g t = v v_{0} = 3.5 μ g t => v = 5 / 7 v_{0} v_{0}^{2} - 25 / 49 v_{0}^{2} = 2 μ g s s = (12 / 49) v_{0}^{2} / μ g$

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The correct option is A 49v20−v2=2as=2μgsv0−v=μgtv=RwTorque=Inertia×αμmgR=0.4mR2α=>α=2.5μg/Rw=αtRw=2.5μgt=vv0=3.5μgt=>v=5/7v0v20−25/49v20=2μgss=(12/49)v20/μg