wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A cart of mass M is at rest on a frictionless horizontal surface and a pendulum bob of mass m hangs from the roof of the cart as given in the figure. The string breaks, the bob falls on the floor, makes several collision on the floor and finally lands up in a small slot made in the floor. The horizontal distance between the string and the slot is L . Find the displacement of the cart during this process.

A
mLM+m towards right.
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
mLM+m towards left
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
MLM+m towards left
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
MLM+M towards right
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B mLM+m towards left
Given that
mass of bob = m
mass of cart = M
horizontal distance between string and slot = L



Initial position of center of mass of the system (from the slot)
=m×L+M×0M+m=mM+mL

Let the bob falls in the slot and M moves by x, as shown

New position of COM is m(x)+M(x)M+m
There is no external force in horizontal direction.
Hence, there is no shift in COM.

m(x)+M(x)M+m=mM+mL
x=mLm+M towards left


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon