Question

A plank of mass 5 kg is placed on a frictionless horizontal plane. Further a block of mass 1 kg is placed over the plank. A massless spring of natural length 2 m is fixed to the plank by its one end. The other end of spring is compressed by the block by half of springs natural length. The system is now released from the rest. The velocity of the plank when block leaves the plank is (the stiffness constant of spring is 100 nm1).

Answer 1

In this question, use conservation of energy principle.

Potential energy stored inthe spring = Kinetic energy of the system afer it is released.

Find the speed then.......................................................................................

Thanks,

Answer 2

Answer:

root(10/3)

Explanation:

Assuming no friction between plank and block.

By momentum conservation ,

⇒1v=5u

⇒v=5u

energy conservation when spring is at neutral position.

⇒  

2

1

​  

kx  

2

=  

2

1

​  

mu  

2

+  

2

1

​  

m  

2

​  

v  

2

 

     

2

1

​  

×100×1  

2

=  

2 1

×5×u  2 +  2

1  ×1×(5u)  

2

 100=30u  

2

⇒u  

=root(10/3)