Question

A marble of mass 2m is travelling with speed 6cm/s and another traveling of mass m with speed 12m/s after collision mass 2m travelling with velocity 9m/s

Answer 1

Answer:

6m/sec

Explanation:

m1*u1+m2*u2=m1*v1+m2*v2

2m*6+m*12=2m*9+m*v2

24m=18m+m*v2

v2=(24m-18m) /m

v2=6m/sec

Answer 2

The final speed of the second marble is 6 m/s.

Explanation:

It is given that,

Mass of the marble is 2m

Initial speed of the marble, u = 6 m/s

Mass of another marble is m

Initial speed of the another marble, u' = 12 m/s

After the collision,

Final speed of the first marble is 9 m/s

We need to find the final speed of the second marble. Let it is v'. It can be calculated using conservation of linear momentum as :

$mu+m'u'=mv+m'v'$

$mu+m'u'-mv=m'v'$

$2m\times 6+m\times 12 -2m\times 9=mv'$    

$v'=\dfrac{6m}{m}$

v' = 6 m/s

So, the final speed of the second marble is 6 m/s. Hence, this is the required solution.

Learn more,

Conservation of momentum

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