Question

A block of mass m-1.0 kg (Block 1) at rest is released at A on a smooth circular surface of radius R-30 m as shown in Figure. At the bottom of the surface another block of mass M 2.0 kg (Block 2) is placed. When Block 1 reaches B, it collides with Block 2 and it stops at B. (20 points) a) What is the potential energy of Block 1 at A in J? (2 points) b) How much kinetic energy does Block 1 have when it reaches 40 30 nu c) What is the velocity of Block 1 just before the collision at B d) Find the velocity of Block 2 right after the collision at B in e) How high does Block 2 reach after the collision in m? (6 points) B before the collision in J? (4 points) in m/s? (2 points) m/s? (6 points)

Answer 1

a) height of block 1 at A=30*(1-cos40)=7.02 m

potential energy at A=1*9.8*7.02=68.8 J

b)kinetic energy at B before colision (from conservation of energy) = 68.8 J

c)let velocity be v

0.5*1*v² = 68.8

v=11.7 m/s

d)from momentum conservation,

1*11.7=2*v

v=5.86 m/s

This is the required answer