Hey there! We receieved your request
Stay Tuned as we are going to contact you within 1 Hour
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-5470-145
+91 7353221155
Use Coupon: CART20 and get 20% off on all online Study Material
Complete Your Registration (Step 2 of 2 )
Sit and relax as our customer representative will contact you within 1 business day
OTP to be sent to Change
A bullet of mass m moving with a velocity v collides head on with a block of mass m which is placed on a car of mass 4m. Bullet gets embedded in the block. The block is connected with car by massless spring of constant k. Find the maximum compression x of the spring. Neglect friction everywhere.
Dear Abdul , Always apply the two principles given below , trying with the force approach maybe complex,
Momentum conservation:
momemtum of bullet = momentum of (bullet + mass m +car)
mv =(m+m+4m) .v2
therefore , v2 = v/6
energy conservation:
k.e of energy of bullet = k.e of ( car +mass+ bullet ) + p.e of spring
1/2 * m* v2 = 1/2*(m+m+4m)*(v/6)2 + 1/2 *k *x2
5mv2/6 = kx2
x = (5mv2/6k)1/2
Please feel free to post as many doubts on our discussion forum as you can. We are all IITians and here to help you in your IIT JEE preparation. Now you can win exciting gifts by answering the questions on Discussion Forum. So help discuss any query on askiitians forum and become an Elite Expert League askiitian.
Win exciting gifts by answering the questions on Discussion Forum.
Sir,
Thanks alot for spending your precious time. My approach is exactly same as yours. But unforunately the answer is not correct.
the correct answer is x=v(m/3k)^1/2
Get your questions answered by the expert for free
You will get reply from our expert in sometime.
We will notify you when Our expert answers your question. To View your Question
Win Gift vouchers upto Rs 500/-
Register Yourself for a FREE Demo Class by Top IITians & Medical Experts Today !