Banner 0
Banner 1
Banner 2
Banner 3
Banner 4
Banner 5
Banner 6
Banner 7
Banner 8
Banner 9

Q.

A solid sphere of mass m = 80 kg and radius r = 0.2 m is released from height h = 5/4 meter. Sphere is initially rotating about horizontal axis passing through its centre of mass. It hits with a stationary cart of mass M = 200 kg exactly at the centre of cart. The cart can move smoothly on the horizontal surface. The collision between sphere and cart occurs in such a way that sphere reaches at same vertical displacement after collision and falls back onto it again. It is found that sphere starts pure rolling at the end of first collision. The coefficient of friction between sphere and cart is  μ=0.1. Match the statement given in List-I to the values given in List-II. (g=10m/sec2)

Question Image
 LIST-I LIST-II
P)The minimum length (in meter) of cart to occur second collision with the sphere1)172
Q)Initial angular velocity 'ω0' (in rad/sec) of sphere on the cart during the process.2)2.8
R)Magnitude of work done (in Joule) by sphere on the cart during the process.3)156
S)Magnitude of work done (in Joule) by cart on the sphere during the process 4)19.5
T) 5)16

 

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya

a

P2;Q4;R5;S1

b

P1;Q3;R1;S3

c

P3;Q5;R4;S2

d

P4;Q3;R4;S1

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

JEE

NEET

NEET

Foundation JEE

Foundation JEE

Foundation NEET

Foundation NEET

CBSE

CBSE

Detailed Solution

v0=2gh=5m/s   …………(i)
Jy=ΔPy=2×80×5=800Ns  …………..(ii)
 Jx=ΔPx
 μJy=mvx

Question Image

vx=1m/s    …………..(iii)
 
  MvC=Jx  …………..(iv)
  vC=0.4m/s …………..(v)
 t0=2×510=1secLmin=2(1+0.4)1=2.8m  …………..(vi)
  At the time of second collision
 Rω'1=0.4
    ω'=7rad/s…………..(vii)
 J'=ΔL
 JxR=I(ω'ω0)
 JxR=I(ω'ω0)
  ω0=19.5rad/sec  ………………(viii)
WmM=12MvC2=16J   ………………(ix)
  WMm=12I(ω')2+12mvx212Iω02=172J

Watch 3-min video & get full concept clarity
score_test_img

courses

No courses found

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Get Expert Academic Guidance – Connect with a Counselor Today!

best study material, now at your finger tips!

  • promsvg

    live classes

  • promsvg

    progress tracking

  • promsvg

    24x7 mentored guidance

  • promsvg

    study plan analysis

download the app

gplay
mentor

Download the App

gplay
whats app icon
personalised 1:1 online tutoring