Question

A small block of mass M=1kg moves on a frictionless surface on an inclined plane. The surface of inclined PQ changes from 60° to 30° at Q. The include of QR is 30°. The block is initially at rest at P. Assuming that there is no loss of energy for collision between the block and the incline(g=10m per s^2), Calculate the speed of block at Q immediately before going on second inclined and the energy at R immediately as it leaves the plane of 30° inclination.

Answer 1

Answer:

Explanation:

Between A and B height fallen by the block is

h1=3–√tan60∘=3m

Hence speed of the block just before striking the second incline is

v2=2gh1−−−−√=2×10×3−−−−−−−−√=60−−√m/s

In perfectly inelastilc collision component of v1 perpendicular to BC will become zero, while component of v1 parallel of BC will remain unchanged.

v2=component of v1 along BC

=v1=cos30∘=(60−−√)(3–√2)=(45−−√)=m/s