Question

a body of mass 20kg teavelling with a velocity 4m/s strikes over another of mass 10kg coming in opposite direction with velicity 6m/s
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Answer 1

Answer:

Let their velocities after the collision be v1 

and v2. As we know for elastic collision. 

Relative velocity of approach = relative velocity of separation

10−4=v2−v1⇒6=v2−v1

⇒v1=v2−6

Applying conservation of momentum,

10×10+5×4=10v1+5v2

120=10v1+5v2

120=10(v2−6)+5v2=15v2−60

15v2=180⇒v2=12cm/sec

v1=6cm/sec