A steel ball of radius R = 20 cm and m = 2 kg is rotating about a horizontal diameter with angular speed ꞷ 0 = 50 rad/sec. This rotating ball is dropped on a rough horizontal floor and fall freely through a height h = 1.25 m. The coefficient of restitution is e = 1 and coefficient of friction between ball and floor is µ = 0.3. Then the distance in m between the point of first and second impact of the ball and floor is
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(3)
Sol. Ball collides with speed V y = 
= 5 m/s and rebound with same speed as e = 1
Impulse of normal reaction = 2mV y = 20 kg s –1 
Impulse of friction = 0.3 × 20 = mV x
or V x = 3 m/s
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems