CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A bullet of mass 200 g fired with a velocity of 200 m/s, hits a block of mass 2 kg moving with a velocity of 10 m/s. After hitting the block, the bullet lodges into it. What is the velocity of the combined system? Assume that the collision is perfectly inelastic.

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

The correct option is A 27.27 m/s
Before collision

After collision

Applying conservation of linear momentum before and after collision.
m1u1+m2u2=(m1+m2)v
0.200×200+2×10=(0.2+2)v
v=27.27 m/s
Velocity of block after collision is 27.27 m/s
The correct option is (a).

flag
Suggest Corrections
thumbs-up
3
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
A Sticky Situation
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon