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  • A ball of mass m, moving with a speed 2v_0 , collides inelastically (e > 0) with an identical ball at rest. Show that (a) For head-on collision, both the balls move forward.

A ball of mass m, moving with a speed 2v_{0} , collides inelastically (e > 0) with an identical ball at rest. Show that
(a) For head-on collision, both the balls move forward.
(b) For a general collision, the angle between the two velocities of scattered balls is less than 90^{\circ}.

Answers (1)

a) We assume, v1 and v2 to be the velocities of the balls post the collision takes place. as we know from the law of conservation of momentum,

MV_{0} = MV_{1} + MV_{2}

From given we can infer that,

2V_{0} = V_{1} + V_{2}----------(1)

E = \frac{V_{2} - V_{1}}{ V_{0} + V_{0}}

Hence, V_{2} = V_{1} + 2EV_{0}

V_{1} = V_{0} (1-e)

Since, e < 1

We can say that the direction of V1 is the same as V0, which is positive and in the forward direction.

b)

The angle between p_{1}and assumed to be θ

From the law of conservation of momentum, \overrightarrow{p}= \overrightarrow{p_{1}} + \overrightarrow{p_{2}}

A portion of kinetic energy is lost as heat when the inelastic collision takes place.

KE(i) > KE_{1} + KE_{2}

\frac{p^{2}}{2m } > \frac{p_{1}^{2}}{2m}+ \frac{p_{2}^{2}}{2m }

hence,\overrightarrow{ p}^{2} > \overrightarrow{p}_{1}^{2} + \overrightarrow{ p}_{2}^{2}

when the angle between p_{1} and p_{2} is less than 90 degrees, the above equation holds.

Posted by

infoexpert24

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