# A small block of mass M moves on a frictionless surface of an inclined plane, as shown in the figure that I have attached with my question the angle of the incline suddenly changes from 60 to 30 degree at point B the block is initially at rest at A assume that collisions between the block and the incline are totally inelastic (g=10m/s²)

1
by harshitarawat80

2014-09-05T14:08:41+05:30

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I assume one needs the find the velocity of the block when it is at the bottom of the inclined plane.

see diagram. the block falls on the lower inclined plane at B. At B the block collides in-elastically (totally). So the component of velocity perpendicular to the lower inclined plane at B becomes zero, as coefficient of restitution is zero. There is loss of energy.

Velocity of block at B is the component of velocity in the direction of inclined plane.

acceleration along inclined plane AB: g cos 30 = (g√3)/2
Distance AB = √[3²+(√3)²] = √12 m

velocity of block at B just before colliding:  V² = u² + 2 a s = 0 + 2 g√3/2 √12
V = √(2 g 3) = √(6g)
= 7.67 m/sec

Velocity just after collision along the lower inclined plane:
V cos 30 = 3√(2g) √3/2 = 3√(3g/2) = 6.64 m/sec

velocity along the lower inclined plane at the bottom of it:
V² = u² + 2 a h

V² = [3√(3g/2)]² + 2 g 3 = 27g/2 + 6 g = 39g/2 = 191.1

V = 13.82 m/sec  along the inclined plane at the bottom of lower inclined plane, just before leaving the plane.

The speed of the block at point c,immediately before it leaves the second incline
Ans this question
both of ur questions are in the answer. hope they r right. check.
Wrong ans
can u tell me the answer values?