A ball X of mass 1 kg travelling at 2 m/s has a head-on collision with an identical ball Y at rest. X stops and Y moves off. Calculate the velocity of Y after the collision.
Here, the mass of the ball X $=m_{X}=1\ kg$
Mass of the ball Y $=m_{Y}=1\ kg$
Before collision:
The velocity of the ball X $=u_{X}=2\ m/s$
The velocity of the ball Y $=u_{Y}=0$
After collision:
The velocity of the ball X $=v_{X}=0$
The velocity of the ball Y $=v_{Y}=?$
According to the law of conservation of momentum:
$m_{X}u_{X}+m_{Y}u_{Y}=m_{X}v_{X}+m_{Y}v_{Y}$
Or $1\ kg\times2\ m/s+1\ kg\times0=1\ kg\times0+1\ kg\times v_{Y}$
Or $v_{Y}=2\ m/s$
Therefore, the velocity of the ball Y is $2\ m/s$.
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