Question

a cube moving on a smooth horizontal surface collides inelastically

Answer 1

Answer:

Consider Initial angular momentum of the block

L

i

=Mv×AC

Angular momentum of the block after it hits the point O,

L

f

=I

c

ω

From parallel axis theorem,

I

o

=I

c

+MxOC

2

=

6

1

Ma

2

+Mx(AC

2

+AO

2

)=

6

1

Ma

2

+

2

1

Ma

2

=

3

2

Ma

2

Using conservation of angular momentum,

2

1

Mva=

3

2

Ma

2

ω

ω=

4a

3v

Answer 2

Answer:

A cube moving on a smooth horizontal surface collides inelastically with a small obstacle which is fixed on the ground. The percentage loss in kinetic energy in this collision is 40%