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 block of mass m is placed on a triangular block of mass M which in turn is placed on a horizontal surface. Assuming frictionless surfaces find the velocity of the triangular block when the smaller block reaches the bottom end.The angle of inclination of the plane is 'theta'.
my approach to the problem........
when the small block reaches the bottom......momentum in x direction is conserved....
mu=Mv....where u and v are absolute velocities of the smallerb block bigger block respectively
applying conservation of energy.....mgh=1/2mu(squared)+1/2mV(squared)
solving the above two equations.....
V=m/M(root of){2Mgh/M+m}
but the answer is complex in terms of angle.....please explain where I am wrong
you have written mgh=1/2mu(square) +1/2Mv(square)
this equation is correct but u(velocity of small block ) should be with respect to ground and here you substituted with respect to bigger block....
so take u=relative velocity of small block with respect to big block
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 !