Question

A marble moving with some velocity collides perfectly elastically head-on with another marble at rest having mass 1.5 times the mass of the colliding marble. The percentage of kinetic energy by the colliding marble after the collision is​

Answer 1

Answer:

Make your co-ordinate axes such that 1 is along the line joining the centres of the 2 just at time on collision and one, tangential. Now consider the velocity of the coming ball in 2 components of the coordinate axes we just assumed, say vn and vt(v-normal and v-tangential). Now during collision, v-tangential can't change as there is no impact along this direction. And collision along normal is just like collision in 1-D. The equations are exactly same. Hence the 1st particle will have 0 velocity along normal and second will move along normal. Hence, they'll move perpendicular to each other.

Answer 2

The kinetic energy by the colliding marble after collision is $\frac{12}{25} mv^{2}$

Given:

Perfectly elastic head on with another mass is 1.5 times

To Find:

The percentage of kinetic energy by the colliding marble after the collision

Solution:

m2 = 1.5  = $\frac{3}{2}$ m1

Conservation of momentum which is just before and after collision is given as,

m1v1 = m1v1' +m2v2'

Conservation of Kinetic energy is given by the below equation,

$\frac{1}{2} m1v1^{2} = \frac{1}{2} m1v1'^{2} + \frac{1}{2} m2v2'^{2}$

Solving these 2 equations we have,

v1' = $\frac{m1-m2}{m1+m2}$

v1 = $\frac{m1 - \frac{3}{2}m1 }{m1+\frac{3}{2} m1} V$

= $\frac{-1}{5}V$

Kinetic energy after collision is ,

= $\frac{1}{2} m(\frac{-1}{5}V )^2$

= $\frac{1}{50} mv^2$

Loss in Kinetic energy = $\frac{1}{2}mv^2$ -  $\frac{1}{50} mv^2$

= $\frac{12}{25} mv^{2}$

The kinetic energy by the colliding marble after collision is   $\frac{12}{25} mv^{2}$      

   #SPJ3