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

A block of mass 2 kg is placed on the top of a bigger block of mass 10 kg as shown in figure. The system is released from rest. Assume all the surfaces are frictionless. Find the distance moved by the bigger block at the instant when the smaller block reaches the ground.


A
8 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
4.5 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1.33 m
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
2.66 m
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C 1.33 m
Take leftward direction as +ve x axis.
Let the distance moved by the bigger block towards right =x m
Then, smaller block will move the distance = (8x) m towards left side
Given, masses m1=2 kg,m2=10 kg

There is no external force on the system (smaller block + bigger block) in the horizontal (x) direction. Since the center of mass of the system was at rest initially, it will remain at rest in the horizontal direction.
i.e change in x - coordinate of COM of the system, Δxcom=0
Δxcom=m1x1+m2x2m1+m2=0

2×(8x)+10(x)2+10=0
162x10x=0
16=12x
x=1.33 m

Distance moved by the bigger block is 1.33 m towards right.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Centre of Mass in Galileo's World
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon