Theory of Intelligence
11TH•Physics•Force And Motion•Inelastic Collision in One Dimension

2.9 Inelastic Collision in One Dimension

Bodies that stick together: conserved momentum and the common final velocity.

2.9 Inelastic Collision in One Dimension

Consider two bodies having masses m1m_{1} and m2m_{2}, moving with velocities v1v_{1} and v2v_{2}, along the same line such that v1>v2v_{1} > v_{2}. In such a case m1m_{1} is regarded as projectile and m2m_{2} as target. After time tt both the bodies make inelastic collision and stick together. Let their combined mass become m1+m2m_{1} + m_{2}, which moves with final velocity vrv_{r} after collision.

Since the collision is perfectly inelastic, the total momentum of balls is conserved. Using law of conservation of momentum:

Total momentum of system before collision = Total momentum of the system after collision

m1v1+m2v2=(m1+m2)vrm_{1}v_{1} + m_{2}v_{2} = (m_{1} + m_{2})v_{r} vr=[m1m1+m2]v1+[m2m1+m2]v2…(2.34)v_{r} = \left[\frac{m_{1}}{m_{1} + m_{2}}\right]v_{1} + \left[\frac{m_{2}}{m_{1} + m_{2}}\right]v_{2} \quad \ldots (2.34)

Which gives the common velocity of the body after inelastic collision.

In a special case when the target m2m_{2} is at rest, v2=0v_{2} = 0, the above equation becomes:

vr=[m1m1+m2]v1v_{r} = \left[\frac{m_{1}}{m_{1} + m_{2}}\right]v_{1}

It shows that velocity of m1m_{1} is reduced by the mass ratio i.e., m1/(m1+m2)m_{1}/(m_{1} + m_{2}).

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