Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • 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 direction of V1 is same as V0, which is positive and in the forward direction.

b)

The angle between p_{1} and p_{2} is 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 lesser than 90 degree, the above equation holds true.

Posted by

infoexpert24

View full answer

Similar Questions

Latest Question