Question

a block of mass 1 kg is moving with a velocity of 10 m/s it collides with a 2kg block which is at has due to collision 1 kg block comes to rest and 2kg block starts moving with a velocity of 4m/s change in momentum of two blocks are respectively:​

Answer 1

Answer:

Step-by-step explanation:

Given :            m  

1

​

=2kg         m  

2

​

=1kg           u  

1

​

=3  m/s        u  

2

​

=4   m/s

Let the common velocity of the combined body be V

Mass of combined body      M=2+1=3kg

Applying conservation of momentum :          

2×3−1×4=3×V              ⟹V=  

3

2

​

   m/s

Momentum of the system     P=2×3−1×4=2 kg m/s

Loss in kinetic energy of the system due to collision         ΔK.E=K.E  

i

​

−K.E  

f

​

 

 ΔK.E=  

2

1

​

(2)×3  

2

+  

2

1

​

(1)×4  

2

−  

2

1

​

(3)×(  

3

2

​

)  

2

 

⟹ΔK.E=  

3

49

​

  J