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

Blocks of masses m, 2m, 4m and 8m are arranged in a line on a frictionless floor. Another block of mass m, moving with speed v along the same line (as shown) collides with mass m in a perfectly inelastic manner. All the subsequent collisions are also perfectly inelastic. By the time the last block of mass 8m starts moving, the total energy loss is p% of the original energy. The value of p is close to-


A
37
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
94
loader
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
77
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
87
loader
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is B 94
According to the question, all collisions are perfectly ineleastic, so after the final collision, all blocks are moving together.


Let the final velocity of the system be v, using momentum conservation,

mv=16mv′
⇒v′=v16

Now initial and final energy of the system is,

Ei=12mv2
Ef=12m(v16)2=12mv216

Energy loss: Ei−Ef=12mv2−12mv216
⇒12mv2[1−116]⇒12mv2[1516]

The total energy loss is p% of the original energy.

∴ p%=Energy lossOriginal energy×100
=12mv2[1516]12mv2×100=93.75%

Hence, value of p is close to 94.

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (B) is the correct answer.

flag
Suggest Corrections
thumbs-up
6
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
The Work Done as a Dot-Product
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon
footer-image